Nguyễn Thanh Hằng

Gọi số học sinh 3 lớp 7A, 7B, 7C lần lượt là a,b , c

Theo bài ta có :

\(\frac{a-10}{7}=\frac{b}{8}=\frac{c+10}{9}\) và \(a+b=85\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\frac{a-10}{7}=\frac{b}{8}=\frac{c+10}{9}=\frac{a-10+b}{7+8}=\frac{75}{15}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{a-10}{7}=5\\\frac{b}{8}=5\\\frac{c+10}{9}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=45\\b=40\\c=35\end{matrix}\right.\)

Vậy...

Ta có :

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)

\(\Leftrightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{147}{\frac{49}{12}}=36\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{\frac{3}{2}}=36\\\frac{y}{\frac{4}{3}}=36\\\frac{z}{\frac{5}{4}}=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=54\\y=48\\z=45\end{matrix}\right.\)

Vậy...

a/ \(4x\left(x-2019\right)-x+2019=0\)

\(\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\)

\(\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2019=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy..

b/ \(3x\left(2x-3\right)=6-4x\)

\(\Leftrightarrow3x\left(2x-3\right)-2\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy..

Ta có :

\(n^3-13n=n^3-n-12n=n\left(n^2-1\right)-12n=n\left(n-1\right)\left(n+1\right)-12n\)

Với mọi số nguyên n ta có :

+) \(n\left(n-1\right)\left(n+1\right)⋮6\) (tích của 3 số nguyên liên tiếp )

+) \(12n⋮6\)

\(\Leftrightarrow n\left(n-1\right)\left(n+1\right)-12n⋮6\)

\(\Leftrightarrow n^3-12n⋮6\left(đpcm\right)\)

\(S=1+7+7^2+......+7^{50}\)

\(\Leftrightarrow7S=7+7^2+......+7^{50}+7^{51}\)

\(\Leftrightarrow7S-S=\left(7+7^2+.....+7^{51}\right)-\left(1+7+....+7^{50}\right)\)

\(\Leftrightarrow6S=7^{51}-1\)

\(\Leftrightarrow S=\frac{7^{51}-1}{6}\)

Ta có :

\(A=n^3-6n^2+9n-2\)

\(=n^3-2n^2-4n^2+8n+n-2\)

\(=n^2\left(n-2\right)-4n\left(n-2\right)+\left(n-2\right)\)

\(=\left(n-2\right)\left(n^2-4n+1\right)\)

Vì A là số nguyên tố

\(\Leftrightarrow\left[{}\begin{matrix}n-2=1\\n^2-4n+1=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=3\\n=0\\n=4\end{matrix}\right.\)

Vậy....

\(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{5}{2}\end{matrix}\right.\)

Vậy...

b/ \(3x^2-7x+4=0\)

\(\Leftrightarrow3x^2-3x-4x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{4}{3}\end{matrix}\right.\)

Vậy...

Ta có :

\(B=x^2+xy+y^2-2x-3y+2019\)

\(\Leftrightarrow4B=4x^2+4xy+4y^2-8x-12y+8076\)

\(\Leftrightarrow4B=\left(4x^2+4xy+y^2\right)-4\left(2x+y\right)+4+3y^2-4y+4022\)

\(\Leftrightarrow2B=\left(2x+y\right)^2-4\left(2x+y\right)+4+3\left(y^2-\frac{4}{3}y+\frac{4}{9}\right)+\frac{12062}{3}\)

\(\Leftrightarrow2B=\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2+\frac{12062}{3}\ge\frac{12062}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{2}{3}\)

Bạn kiểm tra lại nhé, mình k chắc có đúng k nữa !

Bài 2 :

Ta có :

\(\frac{2x-y}{x+y}=\frac{2}{3}\)

\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Leftrightarrow6x-3y=2x+2y\)

\(\Leftrightarrow6x-2x=2y+3y\)

\(\Leftrightarrow4x=5y\)

\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)

Vậy....

Bài 3 : không hiểu đề lắm ???!!!!

Bài 4 :

Ta có :

\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)

Thay (1) ta có :

\(\frac{x}{y}=16\)

\(\Leftrightarrow\frac{2y^2}{y}=16\)

\(\Leftrightarrow2y=16\)

\(\Leftrightarrow y=8\Leftrightarrow x=128\)

Vậy...

Bài 1 :

a/ \(x^2-7x+6=0\)

\(\Leftrightarrow x^2-6x-x+6=0\)

\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

Vậy....

b/ \(x^2-10x+9=0\)

\(\Leftrightarrow x^2-9x-x+9=0\)

\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy...

c/ \(x^2+9x+8=0\)

\(\Leftrightarrow x^2+8x+x+8=0\)

\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)

Vậy ...

d/ \(x^2-11x+10=0\)

\(\Leftrightarrow x^2-11x+10=0\)

\(\Leftrightarrow x^2-x-10x+10=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)

Vậy...

Đặt :

\(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)

\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right)\)

Ta có :

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Lại có : \(\left\{{}\begin{matrix}A\in Z\\1\in Z\end{matrix}\right.\)

\(\Leftrightarrow\frac{4}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\)

Bạn giải các TH ra rồi tự tính x nhé !

quãng đường xe thứ 1 đi đến khi gặp xe 2 là : \(s_1=v_1.t\left(km\right)\)

Quãng đường xe thứ 2 đi đến khi gặp xe 1 là : \(s_2=v_2.t\left(km\right)\)

Khi 2 xe đi ngược chiều :

\(\Leftrightarrow v_1.t+v_2.t=s\)

\(\Leftrightarrow v_1+v_2=105\left(1\right)\)

+) Khi 2 xe đi cùng chiều :

\(v_1.t-v_2.t=s\)

\(\Leftrightarrow3,5\left(v_1-v_2\right)=105\)

\(\Leftrightarrow v_1-v_2=30\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}v_1=67,5km\backslash h\\v_2=37,5km\backslash h\end{matrix}\right.\)

Vậy...

Rđ mắc như thế nào với R1 và R2 thế ? (nối tiếp hay song song)

Áp suất của nước lên đáy thùng là :

\(P=d.h=10000.1,8=18000\left(Pa\right)\)

Đổi \(50cm=0,05m\)

Áp suất của nước lên điểm cách đáy thùng 50cm là :

\(P_A=d.\left(h-h_1\right)=10000\left(1,8-0,05\right)=17500\left(Pa\right)\)

Vậy....

a/ Đặt :

\(\frac{x}{12}=\frac{y}{9}=k\Leftrightarrow\left\{{}\begin{matrix}x=12k\\y=9k\end{matrix}\right.\)

Mà \(xy=12\)

\(\Leftrightarrow12k.9k=12\)

\(\Leftrightarrow k^2=\frac{1}{9}\) \(\Leftrightarrow\left[{}\begin{matrix}k=\frac{1}{3}\\k=-\frac{1}{3}\end{matrix}\right.\)

+) \(k=\frac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

+) \(k=-\frac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-3\end{matrix}\right.\)

Vậy ..

b/ Ta có :

\(\frac{x}{y}=\frac{9}{7}\Leftrightarrow\frac{x}{9}=\frac{y}{7}\left(1\right)\)

\(3y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{3}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{9}=-3\\\frac{y}{7}=-3\\\frac{z}{3}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

Vậy...

Ta có :

\(F=3x-2x^2\)

\(\Leftrightarrow F=-2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)+\frac{9}{8}\)

\(\Leftrightarrow F=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\)

Với mọi x ta có :

\(\left(x-\frac{3}{4}\right)^2\ge0\)

\(\Leftrightarrow-2\left(x-\frac{3}{4}\right)^2\le0\)

\(\Leftrightarrow\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)

\(\Leftrightarrow F\le\frac{9}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{3}{4}\)

Vậy....................

Đề bị sai mk sửa lại :<

Ta có :

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left(a^4-b^4\right)-2ab\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a+b\right)^2\)

\(=\left(a+b\right)^3\left(a-b\right)\)

Ta có :

\(2x+xy=4\)

\(\Leftrightarrow x\left(y+2\right)=4\)

\(\Leftrightarrow y+2=\frac{4}{x}\)

\(\Leftrightarrow y=\frac{4}{x}-2\) \(\left(1\right)\)

Thay (1) vào A ta có :

\(A=x^2y=x^2\left(\frac{4}{x}-2\right)=4x-2x^2\)

\(\Leftrightarrow A=2-2x^2+4x-2\)

\(\Leftrightarrow A=2-2\left(x^2-2x+1\right)\)

\(\Leftrightarrow A=2-2\left(x-1\right)^2\)

Với mọi x ta có :

\(2\left(x-1\right)^2\ge0\)

\(\Leftrightarrow-2\left(x-1\right)^2\le0\)

\(\Leftrightarrow2-2\left(x-1\right)^2\le2\)

\(\Leftrightarrow A\le2\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy..

Ta có :

\(x-\sqrt{x}-6=0\)

\(\Leftrightarrow x+2\sqrt{x}-3\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}\left(x+2\right)-3\left(\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=0\\\sqrt{x}-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(loại\right)\\\sqrt{x}=3\end{matrix}\right.\) \(\Leftrightarrow x=9\)

Vậy...

Ta có :

\(M=a^2+ab+b^2-3a-3b+2001\)

\(\Leftrightarrow2M=2a^2+2ab+2b^2-6a-6b+4002\)

\(\Leftrightarrow2M=\left(a^2+2ab+b^2\right)-4\left(a+b\right)+4+\left(a^2-2a+1\right)+\left(b^2-2a+1\right)+3996\)

\(\Leftrightarrow2M=\left(a+b\right)^2-4\left(a+b\right)+4+\left(a-1\right)^2+\left(b-1\right)^2+3996\)

\(\Leftrightarrow2M=\left(a+b-2\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+3996\)

Với mọi a,b ta có :

\(\left\{{}\begin{matrix}\left(a+b-2\right)^2\ge0\\\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow2M\ge3996\Leftrightarrow M\ge1998\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)

Vậy...

Ta có tứ giác MNPQ là hình bình hành

\(\Leftrightarrow\left\{{}\begin{matrix}MQ=NP\\MQ\backslash\backslash NP\end{matrix}\right.\)

Lại có : E là trung điểm của MQ; F là trung điểm của NP

\(\Leftrightarrow\left\{{}\begin{matrix}EQ=NF\\EQ\backslash\backslash NF\end{matrix}\right.\)

\(\Leftrightarrow ENFQ\:\) là hình bình hành (đpcm)

a/ \(\frac{1}{R_{234}}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{10}+\frac{1}{6}+\frac{1}{9}=\frac{17}{45}\)

\(\Leftrightarrow R_{234}=\frac{45}{17}\left(Ôm\right)\)

\(R_m=R_1+R_{234}=5+\frac{45}{17}=\frac{130}{17}\left(Ôm\right)\)

b/ \(I_m=\frac{U}{R_m}=\frac{15}{\frac{130}{17}}=\frac{51}{26}\left(A\right)=I_1=I_{234}\)

\(U_{234}=I_{234}.R_{234}=\frac{51}{26}.\frac{45}{17}=\frac{135}{26}\left(V\right)=U_2=U_3=U_4\)

\(I_2=\frac{U_2}{R_2}=\frac{\frac{135}{26}}{10}=\frac{27}{52}\left(A\right)\)

\(I_3=\frac{U_3}{R_3}=\frac{\frac{135}{26}}{6}=\frac{45}{52}\left(A\right)\)

\(I_4=\frac{U_4}{R_4}=\frac{\frac{135}{26}}{9}=\frac{15}{26}\left(A\right)\)

Vậy...

a/ \(R_m=R_1+\frac{R_2.R_3}{R_2+R_3}=4+\frac{10.15}{10+15}=10\left(Ôm\right)\)

b/ \(I_m=\frac{U}{R_m}=\frac{5}{10}=\frac{1}{2}\left(A\right)=I_{23}=I_1\)

\(U_{23}=I_{23}.R_{23}=2\left(V\right)=U_2=U_3\)

\(I_2=\frac{U_2}{R_2}=\frac{2}{10}=\frac{1}{5}\left(A\right)\)

\(I_3=\frac{U_3}{R_3}=\frac{2}{15}\left(A\right)\)

Vậy....

\(x^3+6x^2+12x+35=0\)

\(\Leftrightarrow x^3+5x^2+x^2+5x+7x+35=0\)

\(\Leftrightarrow x^2\left(x+5\right)+x\left(x+5\right)+7\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+x+7\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+x+\frac{1}{4}+\frac{27}{4}\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{27}{4}\right]=0\)

\(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)

Vậy...

Ta có :

\(P=2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

\(=2a^2+2b^2-\frac{6a}{b}+\frac{6b}{a}+\frac{9}{a^2}+\frac{9}{b^2}\)

\(=\left(\frac{3}{a^2}+3b^2\right)+\left(\frac{3}{b^2}+3a^2\right)-\left(a^2+2ab+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\left(2ab+\frac{1}{a^2}+\frac{1}{b^2}\right)-10ab\)

\(=\left(\frac{3}{a^2}+3b^2\right)+\left(\frac{3}{b^2}+3a^2\right)-4-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\left(2ab+\frac{1}{a^2}+\frac{1}{b^2}\right)-10ab\)

Áp dụng BĐT Cô si cho các số dương ta có :

\(+,\frac{3}{a^2}+3b^2\ge2\sqrt{\frac{3}{a^2}.3b^2}=\frac{6b}{a}\left(1\right)\)

+, \(\frac{3}{b^2}+3a^2\ge2\sqrt{\frac{3}{b^2}.3a^2}=\frac{6a}{b}\left(2\right)\)

\(+,\left(\frac{a}{b}+\frac{b}{a}\right)\ge2\sqrt{\frac{a}{b}.\frac{a}{b}}=2\Leftrightarrow6\left(\frac{a}{b}+\frac{b}{a}\right)=12\left(3\right)\)

+, \(ab+ab+\frac{1}{a^2}+\frac{1}{b^2}\ge\sqrt{ab.ab.\frac{1}{a^2}.\frac{1}{b^2}}=1\Leftrightarrow6\left(ab+ab+\frac{1}{a^2}+\frac{1}{b^2}\right)=6\)

+) \(ab\ge\frac{\left(a+b\right)^2}{4}\Leftrightarrow10ab\ge10\)

Cộng vế với vế ta có :

\(P\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

\(x^2-\sqrt{x}+x-1\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=\left(\sqrt{x}-1\right)\left(\sqrt{x}+\sqrt{x}+1\right)\)

\(=\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)\)

Vận tốc xuôi dòng của xuồng là :

\(v_{xuôi}=v_x+v_n=20+5=25\left(km\backslash h\right)\)

Vận tốc ngược dòng của xuồng là :

\(v_{ngược}=v_x-v_n=20-5=15\left(km\backslash h\right)\)

Thời gian xuồng đi từ A -> B là :

\(t_1=\frac{s_{AB}}{v_{xuôi}}=\frac{s}{25}\left(h\right)\)

Thời gian xuồng đi từ B -> A là :

\(t_2=\frac{s_{AB}}{v_{ngược}}=\frac{s}{15}\left(km\right)\)

Ta có :

\(t_1+t_2=t\)

\(\Leftrightarrow\frac{s}{25}+\frac{s}{15}=4,8\)

\(\Leftrightarrow s\left(\frac{1}{25}+\frac{1}{15}\right)=4,8\Leftrightarrow s=45\left(km\right)\)

Vậy...

Mạch điện đâu ?!

Đổi \(180s=0,05h\); \(7000m=7km\)

Vận tốc trung bình của xe là :

\(v_{tb}=\frac{s_1+s_2+s_3}{t_1+t_2+t_3}=\frac{2+7+3}{0,05+0,13+4,2}=2,7\left(km\backslash h\right)\)

Vậy...

Ta có :

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x^2+y^2+3^2+2xy+6x+6y\right)+\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(x+y+3\right)^2+\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)

Với mọi y ta có :

\(y^2\ge0\) \(\Leftrightarrow1-y^2\le1\)

\(\Leftrightarrow-1\le x+y+3\le1\)

\(\Leftrightarrow-4\le x+y\le-2\)

\(\Leftrightarrow-6056\le M\le-2019\)

Vậy...

Ta có :

\(x^2=a^2+b^2+ab\)

\(\Leftrightarrow x^4=a^4+3a^2b^2+2a^3b+2ab^3+b^4\)

\(\Leftrightarrow2x^4=2a^4+2b^4+6a^2b^2+4a^3b+4ab^3\)

\(\Leftrightarrow2x^4=a^4+b^4+\left[\left(a^2+2ab+b^2\right)^2\right]\)

\(\Leftrightarrow2x^4=a^4+b^4+\left[\left(a+b\right)^2\right]^2\)

\(\Leftrightarrow2x^4=a^4+b^4+c^4\left(đpcm\right)\)

Đặt :

\(A=1.2+2.3+......+2018.2019\)

\(\Leftrightarrow3A=1.2.3+2.3.3+......+2018.2019.3\)

\(\Leftrightarrow3A=1.2.\left(3-0\right)+2.3\left(4-1\right)+....+2018.2019.\left(2020-2017\right)\)

\(\Leftrightarrow3A=1.2.3-1.2.0+2.3.4-1.2.3+....+2018.2019.2020-2017.2018.2019\)

\(\Leftrightarrow3A=2018.2019.2020\)

\(\Leftrightarrow A=\frac{2018.2019.2020}{3}\)

Vậy....

Ta có :

\(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^5=-c^5\)

\(\Leftrightarrow a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=-c^5\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a^3+b^3+2a^2b+2ab^2\right)\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left[\left(a^3+b^3\right)+2ab\left(a+b\right)\right]\)

\(\Leftrightarrow a^5+b^5+c^5=-5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow a^5+b^5+c^5=-5abc\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow a^5+b^5+c^5\) chia hết cho \(5abc\left(đpcm\right)\)

Ta có :

+, \(\frac{x}{4}=\frac{y}{3}\Leftrightarrow\frac{x}{8}=\frac{y}{6}\left(1\right)\)

+, \(\frac{y}{2}=\frac{z}{5}\Leftrightarrow\frac{y}{6}=\frac{z}{15}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{15}\)

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{8}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{8+6+15}=\frac{5}{29}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{8}=\frac{5}{29}\\\frac{y}{6}=\frac{5}{29}\\\frac{z}{15}=\frac{5}{29}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{40}{29}\\x=\frac{30}{29}\\z=\frac{75}{29}\end{matrix}\right.\)

Vậy...

a/ \(f\left(x\right)=x^3-6x^2+11x-6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

\(=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-5x+6\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

b/ \(f\left(x\right)=x^3-19x-30\)

\(=x^3+3x^2-3x^2-9x-10x-30\)

\(=x^2\left(x+3\right)-3x\left(x+3\right)-10\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x-10\right)\)

\(=\left(x-5\right)\left(x+3\right)\left(x+2\right)\)

c/ \(f\left(x\right)=x^3+4x^2+4x+3\)

\(=x^3+3x^2+x^2+3x+x+3\)

\(=x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+x+1\right)\)

a/ ĐKXĐ : \(x\ge0;x\ne1\)

\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\frac{2}{x^2-2x+1}\)

\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right):\frac{2}{\left(x-1\right)^2}\)

\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\frac{x-2\sqrt{x}+\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-1\right)}{2\left(x-1\right)\left(\sqrt{x}+1\right)}\)

\(=-\sqrt{x}\left(x-1\right)\)

Vậy...

b/ Ta có :

\(P>0\)

\(\Leftrightarrow-\sqrt{x}\left(x-1\right)>0\)

\(\Leftrightarrow\sqrt{x}\left(x-1\right)< 0\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ

Vậy \(0< x< 1\) thì P > 0

c/ Ta có :

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) thỏa mãn \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thay vào P rồi bạn tự tính ra nhé :>

a/ ĐKXĐ : \(x\ge0;x\ne9;x\ne4\)

Ta có :

\(P=\left(\frac{2\sqrt{x}}{9-x}+\frac{1}{3+\sqrt{x}}\right).\frac{x\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\)

\(=\left(\frac{2\sqrt{x}}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{3-\sqrt{x}}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right).\frac{x\left(3-\sqrt{x}\right)}{\sqrt{x-2}}\)

\(=\frac{\sqrt{x}+3}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\frac{x\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\)

\(=\frac{1}{\sqrt{x}-2}\)

Vậy \(P=\frac{1}{\sqrt{x}-2}\) với ĐKXĐ \(x\ge0;x\ne9;x\ne4\)

b/ Với ĐKXĐ \(x\ne0;x\ne9;x\ne4\) ta có :

\(P=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-2}=-\frac{1}{3}\)

\(\Leftrightarrow2-\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=-1\) (vô lí)

Vậy không tìm đc x thỏa mãn

Gọi quãng đường từ ga đến trường là \(s\left(s>0\right)\)

Ta có :

\(s=v_1.t_1=v_2.t_2\)

\(\Leftrightarrow12t=5\left(t+0,25\right)\)

\(\Leftrightarrow12t-5t=\frac{5}{4}\)

\(\Leftrightarrow7t=\frac{5}{4}\)

\(\Leftrightarrow t=\frac{5}{28}\)

Vậy quãng đường là : \(s=v_1.t_1=12t=\frac{15}{7}\left(km\right)\)

Vậy...

Ta có :

\(A=4x^3+11x^2+5x+5\)

\(=4x^3+8x^2+3x^2+6x-5x-10+15\)

\(=4x^2\left(x+2\right)+3x\left(x+2\right)-5\left(x+2\right)+15\)

\(=\left(x+2\right)\left(4x^2+3x-5\right)+15\)

\(\Leftrightarrow A\) chia hết cho \(x+2\) thì \(x+2\inƯ\left(15\right)\)

Ta có các TH :

+, \(x+2=1\Leftrightarrow x=-1\)

+, \(x+2=-1\Leftrightarrow x=-3\)

+, \(x+2=15\Leftrightarrow x=13\)

+, \(x+2=-15\Leftrightarrow x=-17\)

+, \(x+2=-3\Leftrightarrow x=-5\)

+, \(x+2=3\Leftrightarrow x=1\)

+, \(x+3=5\Leftrightarrow x=2\)

+, \(x+3=-5\Leftrightarrow x=-8\)

Vậy...