Nguyễn Thanh Hằng

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne5\\x\ne\pm15\end{matrix}\right.\)

\(\frac{1}{5-x}=\frac{4}{15+x}+\frac{4}{15-x}\)

\(\Leftrightarrow\frac{1}{5-x}=\frac{4\left(15-x\right)}{\left(15+x\right)\left(15-x\right)}+\frac{4\left(15+x\right)}{\left(15+x\right)\left(15-x\right)}\)

\(\Leftrightarrow\frac{1}{5-x}=\frac{60-4x+60+4x}{\left(15+x\right)\left(15-x\right)}\)

\(\Leftrightarrow\frac{1}{5-x}=\frac{120}{\left(15+x\right)\left(15-x\right)}\)

\(\Leftrightarrow\left(15+x\right)\left(15-x\right)=120\left(5-x\right)\)

\(\Leftrightarrow225-x^2=600-120x\)

\(\Leftrightarrow x^2-120x+375=0\)

Ta có : \(\Delta'=60^2-375=3225>0\)

Khi đó phương trình có 2 nghiệm phân biệt là :

\(x_1=\frac{b'-\sqrt{\Delta'}}{a}=60-\sqrt{3225}=6-5\sqrt{129}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=-60-\sqrt{3225}=-60-5\sqrt{129}\)

Vậy....

a/ ĐKXĐ : \(\left\{{}\begin{matrix}z\ne0\\z\ne4,-4\end{matrix}\right.\)

Thay \(z=1\) vào biểu thức A ta có :

\(A=\frac{2.1}{1^2-4}=\frac{2}{-3}=-\frac{2}{3}\)

Vậy....

b/ Ta có :

\(B=\frac{2z^3+4z}{z^4-8z^2+16}\)

\(=\frac{2z\left(z+2\right)}{\left(z^2-4\right)^2}\)

\(=\frac{2z\left(z+2\right)}{\left(z-2\right)^2\left(z+2\right)^2}\)

\(=\frac{2z}{\left(z-2\right)^2\left(z+2\right)}\)

Lại có : \(M=A:B\)

\(\Leftrightarrow M=\frac{2z}{\left(z-2\right)\left(z+2\right)}:\frac{2z}{\left(z-2\right)^2\left(z+2\right)}\)

\(=\frac{2z}{\left(z-2\right)\left(z+2\right)}.\frac{\left(z-2\right)^2\left(z+2\right)}{2z}\)

\(=z-2\)

Vậy...

1.

a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)

\(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left[\left(x+y\right)^2-2xy\right]\)

\(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\)

\(=2-6xy-3+6xy\)

\(=-1\)

Vậy...

2.

a. \(127^2+146.127+73^2\)

\(=127^2+2.73.127+73^2\)

\(=\left(127+73\right)^2=200^2=40000\)

b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1\)

\(=1\)

Đặt :

\(\frac{a}{b}=\frac{c}{d}=k\)

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

Ta có :

+) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(1\right)\)

+) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\frac{b^2}{d^2}\left(2\right)\)

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

Ta có :

\(x< -3\Leftrightarrow x+3< 0\)

Lại có :

\(3x+\sqrt{9+6x+x^2}\)

\(=3x+\sqrt{\left(x+3\right)^2}\)

\(=3x+\left|x+3\right|\)

\(=3x-x-3\) (Vì \(x+3< 0\) )

\(=2x-3\)

Vậy...

\(x-4\sqrt{x^2+3}=0\)

\(\Leftrightarrow x=4\sqrt{x^2+3}\)

\(\Leftrightarrow x^2=\left(4\sqrt{x^2+3}\right)^2\)

\(\Leftrightarrow x^2=16\left(x^2+3\right)\)

\(\Leftrightarrow x^2=16x^2+48\)

\(\Leftrightarrow15x^2+48=0\)

\(\Leftrightarrow15x^2=-48\) (vô lí)

Vậy...

a/ \(ax-2x-a^2+2a\)

\(=x\left(a-2\right)-a\left(a-2\right)\)

\(=\left(x-a\right)\left(a-2\right)\)

Vậy....

b/ \(x^2+x-ax-a\)

\(=x\left(x+1\right)-a\left(x+1\right)\)

\(=\left(x+1\right)\left(x-a\right)\)

Vậy...

d/ \(2xy-ax+x^2-2ay\)

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

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

e/ \(x^3+ax^2+x+a\)

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

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

Vậy...

a/ \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)

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

\(=3\)

Vậy......

b/ \(x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)

\(=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x\)

\(=16\)

Vậy..

a/ \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

b/ \(x^2-22=\left(x-\sqrt{22}\right)\left(x+\sqrt{22}\right)\)

đăng từng bài một lên thui bạn ơi ;'<

a/ \(x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)

b/ \(a^2+2ab+b^2-ac-bc=\left(a^2+2ab+b^2\right)-\left(ac+bc\right)=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

c/ \(25x^2-10x-3x=25x^2-13x=x\left(25x-13\right)\)

d/ \(16x^2+24x-7=16x^2-4x+28x-7=4x\left(4x-1\right)+7\left(4x-1\right)=\left(4x-1\right)\left(4x+7\right)\)

ĐKXĐ : \(x\ne\mp y\) ; \(x,y\ne0\)

Ta có :

\(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2+y^2}\right):\frac{4xy}{y^2-x^2}\)

\(=\left(\frac{1}{\left(x+y\right)^2}-\frac{1}{\left(x-y\right)\left(x+y\right)}\right):\frac{4xy}{\left(y-x\right)\left(x+y\right)}\)

\(=\left(\frac{x-y}{\left(x-y\right)\left(x+y\right)^2}-\frac{x+y}{\left(x-y\right)\left(x+y\right)^2}\right).\frac{\left(y-x\right)\left(x+y\right)}{4xy}\)

\(=\frac{x-y-x-y}{\left(x-y\right)\left(x+y\right)^2}.\frac{\left(y-x\right)\left(x+y\right)}{4xy}\)

\(=\frac{-2y}{\left(x-y\right)\left(x+y\right)^2}.\frac{\left(y-x\right)\left(x+y\right)}{4xy}\)

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

Vậy..

Xét \(\Delta ABC\) vuông tại A

\(BC^2=AB^2+AC^2\) (Định lí Py ta go)

\(\Leftrightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+7,5^2}=8,5cm\)

Xét \(\Delta ABC\) vuông tại A
Đường cao AH

\(\Leftrightarrow AB^2=BH.BC\)

\(\Leftrightarrow BH=\frac{AB^2}{BC}=\frac{4^2}{8,5}=\frac{32}{17}cm\)

\(\Leftrightarrow CH=BC-BH=8,5-\frac{32}{17}=\frac{225}{17}cm\)

Vậy..

a/ \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|+\left|2\sqrt{3}-\sqrt{2}\right|=\sqrt{3}+\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}\) (Vì \(\sqrt{3}+\sqrt{2}>0\) , \(2\sqrt{3}-\sqrt{2}>0\) )

b/ \(\sqrt{5-2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}-\sqrt{2}\right|+\left|2\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}-2\sqrt{2}\)

c/ \(11-\sqrt{33}=\sqrt{11}.\sqrt{11}-\sqrt{3}.\sqrt{11}=\sqrt{11}\left(\sqrt{11}-\sqrt{3}\right)\)

ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

Ta có :

\(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

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

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

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

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

\(=1\)

Vậy...

1/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Thay \(x=9\) vào biểu thức A ta có :

\(A=\frac{\sqrt{9}+1}{\sqrt{9}-1}=\frac{3+1}{3-1}=2\)

Vậy...

2/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có :

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

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

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

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)

Vậy....

b/ Ta có :

\(2P=2\sqrt{x}+5\)

\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\)

\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)

\(\Leftrightarrow2\sqrt{x}-1=0\)

\(\Leftrightarrow x=\frac{1}{4}\)

Vậy..

b/ Ta có :

\(M=\frac{3^2}{2.5}+\frac{3^2}{5.8}+....+\frac{3^2}{98.101}\)

\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{98.101}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{98}-\frac{1}{101}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=3.\frac{99}{202}\)

\(=\frac{297}{202}\)

Vậy....

1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))

\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)

2/Ta có :

\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\sqrt{3}-3}-\frac{5\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\left(\sqrt{3}-1\right)}-\frac{5\sqrt{6}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}\right).\frac{1}{\sqrt{6}}\)

\(=\frac{3\sqrt{2}-15\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-7+\sqrt{3}}{6}\)

Vậy...

a/ \(\frac{2x^3}{4x^7}=\frac{1}{2x^4}\) với ĐKXĐ : \(x\ne0\)

b/ \(\frac{x-1}{\left(x+1\right)^2}.\frac{x^2+2x+1}{x^2-1}=\frac{x-1}{\left(x+1\right)^2}.\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x+1}\) với ĐKXĐ : \(x\ne\pm1\)

c/ \(\frac{x^2-7x+12}{x^2-16}=\frac{\left(x-4\right)\left(x-3\right)}{\left(x-4\right)\left(x+4\right)}=\frac{x-3}{x+4}\) với ĐKXĐ : \(x\ne\pm4\)

d/ \(\frac{x-1}{\sqrt{x}+1}:\left(\sqrt{x}-1\right)=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=1\) với ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Hình bạn tự vẽ nhé <3

a/ Xét tam giác ABD vuông tại A

\(\Leftrightarrow BD^2=AB^2+AD^2\) (Định lí Py ta go)

\(\Leftrightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{8^2+15^2}=17\)

Vậy....

b/ Xét tam giác ABD vuông tại A

Đường cao AH

\(\Leftrightarrow BD.AH=AB.AD\)

\(\Leftrightarrow AH=\frac{AB.AD}{BD}=\frac{8.15}{17}=\frac{120}{17}\)

Vậy....

a/ Ta có :

\(BH+CH=BC\) (H nằm giữa B và C)

Lại có : \(CH=4BH\)

\(\Rightarrow BC=5BH\)

Xét \(\Delta ABC\) vuông tại A

Đường cao AH

\(\Leftrightarrow AB^2=BH.BC\)

Mà \(AH=2\sqrt{5};BC=5BH\)

\(\Leftrightarrow\left(2\sqrt{5}\right)^2=BH.5BH\)

\(\Leftrightarrow BH^2=4\)

\(\Leftrightarrow BH=2\)

\(\Leftrightarrow CH=4.BH=8\)

\(\Leftrightarrow BC=BH+CH=2+9=10\)

b/ Xét tam giác ABC vuông tại A

\(BC^2=AB^2+AC^2\) (định lí Py ta go)

\(\Leftrightarrow AC=\sqrt{10^2-\left(2\sqrt{5}\right)^2}=4\sqrt{5}\)

Ta có :

\(S_{ABC}=\frac{1}{2}AB.AC=\frac{1}{2}.2\sqrt{5}.4\sqrt{5}=20\)

Vậy..

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{5}=\frac{b}{3}=\frac{c}{10}=\frac{d}{7}=\frac{a+b+c+d}{5+3+10+7}=\frac{125}{25}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{a}{5}=5\\\frac{b}{3}=5\\\frac{c}{10}=5\\\frac{d}{7}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=25\\b=15\\c=50\\d=35\end{matrix}\right.\)

Vậy.....

Ta có :

\(a:b:c:d=2:5:8:15\)

\(\Leftrightarrow\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{d}{25}\)

và \(a+b+c+d=90\)

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{d}{15}=\frac{a+b+c+d}{2+5+8+15}=\frac{90}{30}=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{a}{2}=3\\\frac{b}{5}=3\\\frac{c}{8}=3\\\frac{d}{15}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=15\\c=24\\d=45\end{matrix}\right.\)

Vậy...

Hình bạn tự vẽ nhé <3

a/ Xét \(\left(O,R\right)\) có :

\(\left\{{}\begin{matrix}\widehat{AMH}=90^0\\\widehat{ANH}=90^0\end{matrix}\right.\) (góc nội tiếp chắn nửa đường tròn)

Xét \(\Delta AHB\) vuông tại H

Đường cao HM

\(\Leftrightarrow AH^2=AM.AB\left(1\right)\)

Xét \(\Delta AHC\) vuông tại H

Đường cao HN

\(\Leftrightarrow AH^2=AN.AC\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow AM.AB=AN.AC\left(đpcm\right)\)

b/ Ta có :

\(\left\{{}\begin{matrix}\widehat{HAC}+\widehat{ACH}=90^0\\\widehat{AMN}+\widehat{NMH}=90^0\end{matrix}\right.\)

Mà \(\widehat{NMH}=\widehat{NAH}\) (góc nội tiếp cùng chắn cung NH)

\(\Leftrightarrow\widehat{ACH}=\widehat{NMH}\)

Xét tứ giác \(BMNH\) có :

\(\widehat{ACH}=\widehat{NMH}\)

\(\Leftrightarrow\) Tứ giác BMNH nội tiếp

Ta có :

\(\sqrt{4\left(1-x\right)^2}-6=0\)

\(\Leftrightarrow\sqrt{4\left(1-x\right)^2}=6\)

\(\Leftrightarrow4\left(1-x\right)^2=36\)

\(\Leftrightarrow\left(1-x\right)^2=9\)

\(\Leftrightarrow1-2x+x^2=9\)

\(\Leftrightarrow x^2-2x+1-9=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

Vậy....

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

\(\Leftrightarrow3x^2-3x-2x^2-4x=-1-x\)

\(\Leftrightarrow3x^2-2x^2-3x-4x+x+1=0\)

\(\Leftrightarrow x^2-6x+1=0\)

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

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

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

Vậy....

Vận tốc của An : \(v_1=10km\backslash h\)

Vận tốc của Bình : \(v_2=170m\backslash p=\frac{170:1000}{1:60}=10,2km\backslash h\)

Vận tốc của Chi : \(v_3=2,8m\backslash s=\frac{2,8:1000}{1:3600}=10,08km\backslash h\)

\(\Leftrightarrow\) Bình là người chạy nhanh nhất

Ta có : \(B=\frac{1+2+2^2+....+2^{2008}}{1-2^{2009}}\)

Đặt :

\(H=1+2+2^2+2^3+.........+2^{2008}\) \(\Leftrightarrow B=\frac{H}{1-2^{2009}}\)

\(\Leftrightarrow2H=2+2^2+2^3+......+2^{2008}+2^{2009}\)

\(\Leftrightarrow2H-H=\left(2+2^2+.....+2^{2009}\right)-\left(1+2+2^2+....+2^{2008}\right)\)

\(\Leftrightarrow H=2^{2009}-1\)

\(\Leftrightarrow B=\frac{2^{2009}-1}{1-2^{2009}}=\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

Vậy...

Áp dụng BĐT Bunhiacopxki ta có :

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

\(\Leftrightarrow C\le\sqrt{2.2}\)

\(\Leftrightarrow C\le2\)

Dấu "=" xảy ra \(\Leftrightarrow x=2-x\)

\(\Leftrightarrow x=1\)

Vậy...

Ta có :

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}< 1\)

\(\Leftrightarrow M< 1\)

Vậy....

Áp dụng BĐT Bunhiacopxki ta có :

\(B=\sqrt{6-x}+\sqrt{x+2}\le\sqrt{\left(1+1\right)\left(6-x+x+2\right)}\)

\(\Leftrightarrow B\le\sqrt{2.8}=4\)

Dấu "=" xảy ra \(\Leftrightarrow6-x=x+2\)

\(\Leftrightarrow x=2\)

Vậy...

a/ \(44a^2-44a+11\)

\(=11\left(4a^2-4a+1\right)\)

\(=11\left(2a-1\right)^2\)

Vậy...

b/ \(2a^3+3a^2+3a+1\)

\(=2a^3+a^2+2a^2+a+2a+1\)

\(=a^2\left(2a+1\right)+a\left(2a+1\right)+\left(2a+1\right)\)

\(=\left(2a+1\right)\left(a^2+a+1\right)\)

Vậy..

Hình bạn tự vẽ nhé <3

a/ Ta có : \(\left\{{}\begin{matrix}AB+AE=AE\\BD+DC=BC\end{matrix}\right.\)

Mà \(\left\{{}\begin{matrix}AB=BD\\AE=DC\end{matrix}\right.\)

\(\Leftrightarrow BE=BC\)

Xét \(\Delta EBD\) và \(\Delta CBA\) có :

\(\left\{{}\begin{matrix}\widehat{EBC}chung\\BD=AB\\BE=BC\end{matrix}\right.\)

\(\Leftrightarrow\Delta EBD=\Delta CBA\left(c-g-c\right)\)

\(\Leftrightarrow\widehat{BED}=\widehat{ÂCB}\left(đpcm\right)\)

b/ \(\Delta EBD=\Delta CBA\left(cmt\right)\)

\(\Leftrightarrow\) \(\widehat{BAC}=\widehat{BDE}\)

Lại có : \(\left\{{}\begin{matrix}\widehat{BAC}+\widehat{EAI}=180^0\\\widehat{BDE}+\widehat{IDC}=180^0\end{matrix}\right.\)

\(\Leftrightarrow\widehat{EAI}=\widehat{IDC}\)

Xét \(\Delta AIE\) và \(\Delta DIC\) có :

\(\left\{{}\begin{matrix}\widehat{AEI}=\widehat{ICD}\\AE=DC\\\widehat{IAE}=\widehat{IDC}\end{matrix}\right.\)

\(\Leftrightarrow\Delta AIE=\Delta DIC\left(g-c-g\right)\)

\(\Leftrightarrow IE=IC\)

c/ Xét \(\Delta BEI\) và \(\Delta BCI\) có :

\(\left\{{}\begin{matrix}BE=BC\\IE=IC\\\widehat{BED}=\widehat{ACB}\end{matrix}\right.\)

\(\Leftrightarrow\Delta BEI=\Delta BCI\left(c-g-c\right)\)

\(\Leftrightarrow\widehat{EBI}=\widehat{IBC}\)

\(\Leftrightarrow BI\) là tia phân giác của \(\widehat{EBC}\left(đpcm\right)\)

Hình bạn tự vẽ nhé <3

a/ Xét \(\Delta ABD\) và \(\Delta EBD\) có :

\(\left\{{}\begin{matrix}\widehat{BAD}=\widehat{BED}=90^0\\BDchung\\\widehat{ABD}=\widehat{DBE}\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABD=\Delta EBD\left(ch-gn\right)\)

\(\Leftrightarrow AB=EB\left(đpcm\right)\)

b/ Xét \(\Delta ABC\) và \(\Delta EBF\) có :

\(\left\{{}\begin{matrix}\widehat{ABC}=\widehat{BEF}=90^0\\AB=EB\\\widehat{ABC}chung\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABC=\Delta EBF\left(g-c-g\right)\)

c/ \(AB=BE\)

\(\Leftrightarrow\Delta ABE\) cân tại B

Lại có : BD là tia phân giác của \(\widehat{ABE}\)

\(\Leftrightarrow BD\) là đường trung trực của \(AE\left(đpcm\right)\)

d/ \(\Delta ABC=\Delta EBF\left(cmt\right)\)

\(\Leftrightarrow BF=BC\)

\(\Leftrightarrow\Delta FBC\) cân tại B

Lại có : M là trung điểm của FC

\(\Leftrightarrow BM\) là tia phân giác của \(\widehat{FBC}\)

Mà BD cũng là tia phân giác của \(\widehat{FBC}\)

\(\Leftrightarrow\) 3 điểm \(B,D,M\) thẳng hàng \(\left(đpcm\right)\)

a/ \(x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\)

\(=xy-xz+yz-xy+zx-yz\)

\(=0\)

Vậy...

b/ \(x\left(y+z-yz\right)-y\left(z+x-zx\right)+zy+x\)

\(=xy+xz-xyz-yz-xy+xyz+zy+x\)

\(=x\)

Vậy....

a/ \(\sqrt{5+\sqrt{24}}-\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{2}=\left|\sqrt{3}+\sqrt{2}\right|-\sqrt{2}=\sqrt{3}+\sqrt{2}-\sqrt{2}=\sqrt{3}\)

b/ \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{3}\)

c/ \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)

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

Hình bạn tự vẽ nhé <3

a/ Ta có :

+) \(\widehat{ABH}+\widehat{BAH}=90^0\) (2 góc phụ nhau)

+) \(\widehat{BAH}+\widehat{IAC}=90^0\) (2 góc phụ nhau)

\(\Leftrightarrow\widehat{ABH}=\widehat{IAC}\left(đpcm\right)\)

b/ Xét \(\Delta ABH\) và \(\Delta CAI\) có :

\(\left\{{}\begin{matrix}\widehat{ABH}=\widehat{IAC}\\AB=AC\\\widehat{AHB}=\widehat{AIC}=90^0\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABH=\Delta CAI\left(g-c-g\right)\)

\(\Leftrightarrow AH=CI\) \(\left(đpcm\right)\)

c/ Xét \(\Delta ABD\) có :

\(\left\{{}\begin{matrix}AM\perp BD\\BH\perp AD\end{matrix}\right.\)

Lại có :

K là giao điểm của AM và BH

\(\Leftrightarrow K\) là trực tâm của \(\Delta ABD\)

\(\Leftrightarrow DK\perp AB\left(đpcm\right)\)

Xét đa thức :

\(f\left(x\right)=ax^2+bx+c\)

Ta có :

+) \(f\left(0\right)=1\)

\(\Leftrightarrow a.0^2+b.0+c=1\)

\(\Leftrightarrow c=1\)

+) \(f\left(1\right)=-1\)

\(\Leftrightarrow a.1^2+b.1+c=-1\)

\(\Leftrightarrow a+b+c=-1\)

\(\Leftrightarrow a+b=-2\)

Vậy..