Nguyễn Thanh Hằng

mạch đâu ?

Bài 1 :

a/ Trọng lực của vật là :

\(P=10m=10.60=600\left(N\right)\)

Công có ích là :

\(A_i=P.h=600.1,5=900\left(J\right)\)

b/ Công toàn phần là :

\(A_{tp}=F.l=100.10=1000\left(J\right)\)

Hiệu suất của mặt phẳng nghiêng là :

\(H=\frac{A_i}{A_{tp}}.100\%=\frac{900}{1000}.100\%=90\%\)

Vậy..

Ta có :

+) \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Leftrightarrow\)\(\frac{ayz+bxz+cxy}{xyz}=0\) \(\left(1\right)\)

Lại có :

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

\(\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=0\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)

Lần sau bạn gõ công thức nhé !

Ta có :

\(A=\left|3x-3\right|-111\)

Với mọi x ta có :

\(\left|3x-3\right|\ge0\)

\(\Leftrightarrow\left|3x-3\right|-111\ge-111\)

\(\Leftrightarrow A\ge-111\)

Dấu "=" xảy ra \(\Leftrightarrow\left|3x-3\right|=0\Leftrightarrow x=-1\)

Vậy \(A_{Min}=-111\Leftrightarrow x=1\)

2.

+) Nếu \(x+y+z+t=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\z+t=-\left(y+z\right)\\y+z=-\left(x+t\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)

Ta có :

\(M=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)

\(=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(x+t\right)}{x+t}+\frac{-\left(y+z\right)}{y+z}+\frac{-\left(y+z\right)}{y+z}\)

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

\(=-4\)

+) Nếu \(x+y+z+t\ne0\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3\left(x+y+z+t\right)}=\frac{1}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=y+z+t\\3y=x+z+t\\3z=x+y+t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=x+y+z+t\\4y=x+y+z+t\\4z=x+y+z+t\end{matrix}\right.\)

\(\Leftrightarrow x=y=z=t\)

Ta có :

\(M=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)

\(=\frac{2x}{2x}+\frac{2x}{2x}+\frac{2x}{2x}+\frac{2x}{2x}\)

\(=1+1+1+1=4\)

Vậy..

Đặt :

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+........+\frac{100}{3^{100}}\)

\(\Leftrightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{100}{3^{100}}\right)\)

\(\Leftrightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt : \(H=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)

\(\Leftrightarrow3H=3+1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(\Leftrightarrow3H-H=\left(4+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow2H=3-\frac{1}{3^{99}}\)

\(\Leftrightarrow H=\frac{3-\frac{1}{99}}{2}\)

\(\Leftrightarrow2A=\frac{3-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{2.3^{100}}\)

\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)

\(\left(x-5\right)^{x+3}-\left(x-5\right)^{x+13}=0\)

\(\Leftrightarrow\left(x-5\right)^{x+3}.\left[1-\left(x-5\right)^{x+10}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{x+3}=0\\1-\left(x-5\right)^{x+10}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{x+10}=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

Vậy..

a/ Thể tích thỏi chì là :

\(V=380-180=200\left(cm^3\right)=0,0002m^3\)

b/ Khối lượng thỏi chì là :

\(m=D.V=0,0002.11300=2,26\left(kg\right)\)

Trọng lượng thỏi chì là :

\(P=10m=2,26.10=22,6\left(N\right)\)

c/ Lực kéo đó nhỏ hơn trọng lượng của thỏi chì .

Đầu máy phải tác dụng vào xe một lực kéo là :

\(F=\frac{A}{s}=\frac{120000}{30}=4000\left(N\right)\)

Vậy...

a/ Khối lượng riêng của vật là :

\(D=\frac{m}{V}=\frac{180}{1,2}=150\left(km\backslash m^3\right)\)

Trọng lượng riêng của vật là :

\(d=10.D=150.10=1500\left(N\backslash m^3\right)\)

Vậy...

b/ Trọng lượng của vật là : \(P=10m=180.10=1800\left(N\right)\)

Vậy..

Vật chìm hoàn toàn \(\Leftrightarrow V_v=V_c\)

Thể tích vật chìm trong nước là :

\(V_c=a.b.c=20.15.20=6000\left(cm^3\right)=0,006\left(m^3\right)\)

Lực đẩy Ác - si - mét tác dụng lên vật là :

\(F_A=d.V_c=10000.0,006=60\left(N\right)\)

Vậy..

\(n_C=\frac{m}{M}=\frac{24}{12}=2\left(mol\right)\)

\(n_{O_2}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH :

\(C+O_2\rightarrow CO_2\)

Ta có : \(n_C:n_{O_2}=\frac{2}{1}:\frac{0,5}{1}=2:0,5\)

\(\Leftrightarrow C\) dư , \(O_2\) hết

\(n_Cpư=n_{O_2}=0,5\left(mol\right)\)

\(\Leftrightarrow n_C\) dư \(=2-0,5=1,5\left(mol\right)\)

\(\Leftrightarrow m_Cdư=n.M=12.1,5=18\left(g\right)\)

b/ \(n_{CO2}=n_{O2}=0,5\left(mol\right)\)

\(\Leftrightarrow m_{CO2}=n.M=0,5.44=22\left(g\right)\)

Vậy...

Ta có :

\(\left|x-1\right|=\left|1-x\right|\)

\(\Leftrightarrow\left|x-1\right|+\left|x+3\right|=\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|\)

\(\Leftrightarrow\left|x-1\right|+\left|x+3\right|\ge\left|4\right|\)

\(\Leftrightarrow\left|x-1\right|+\left|x+3\right|\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(1-x\right)\left(x+3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-x\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x\le0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1\ge x\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}1\le x\\x\le-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1\ge x\ge-3\\x\in\varnothing\end{matrix}\right.\)

Vậy..

5. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{x+10}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{x+10}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{x+10}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

Vậy ...

4/ Ta có :

\(3^{n+2}-2^{n+2}+3^n-2^n=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)

3/ Ta có :

\(A=1+5+5^2+......+5^{49}+5^{50}\)

\(\Leftrightarrow5A=5+5^2+5^3+.......+5^{50}+5^{51}\)

\(\Leftrightarrow5A-A=\left(5+5^2+.........+5^{51}\right)-\left(1+5+5^2+....+5^{50}\right)\)

\(\Leftrightarrow4A=5^{51}-1\)

\(\Leftrightarrow A=\frac{5^{51}-1}{4}\)

Vậy..

a. PTHH :

\(4Al+3O_2\rightarrow2Al_2O_3\)

b/ \(n_{O_2}=\frac{V}{22,4}=\frac{28}{22,4}=1,25\left(mol\right)\)

Theo pt : \(n_{Al}=\frac{4}{3}.n_{O_2}=1,25.\frac{4}{3}=\frac{5}{3}\left(mol\right)\)

\(\Leftrightarrow m_{Al}=m.M=\frac{5}{3}.27=45\left(g\right)\)

c/ \(n_{Al2O3}=\frac{2}{3}.n_{O_2}=\frac{2}{3}.1,25=\frac{2,5}{3}\left(mol\right)\)

\(\Leftrightarrow m_{Al_2O_3}=n.M=\frac{2,5}{3}.102=85\left(g\right)\)

Vậy...

Ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)

\(\Leftrightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=1\)

\(\Leftrightarrow\frac{a+b+c}{abc}=1\Leftrightarrow a+b+c=abc\left(đpcm\right)\)

PTHH : \(4Al+3O_2\rightarrow2Al_2O_3\)

Theo định luật bảo toàn khối lượng ta có :

\(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

\(\Leftrightarrow m_{O_2}=8-5=3\left(g\right)\)

Vậy...

Look those dark clouds ! it....in some minutes .

A. is raining B. rains C. is going to rain D. will Rain

Ta có :

\(\left|x-5\right|-x=3\)

\(\Leftrightarrow\left|x-5\right|=x+3\)

Lại có : \(\left|x-5\right|\ge0\forall x\)

\(\Leftrightarrow x+3\ge0\Leftrightarrow x\ge-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x+3\\x-5=-x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-x=5+3\\x+x=-3+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=8\left(vôli\right)\\2x=2\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy.....

PTHH :

\(H_2+Cl_2\rightarrow2HCl\left(1\right)\)

\(HCl+AgNO_3\rightarrow AgCl+HNO_3\) \(\left(2\right)\)

Ta có : \(n_{H_2}=\frac{V}{22,4}=\frac{1}{22,4}=0,446\left(mol\right)\)

\(n_{Cl_2}=\frac{V}{22,4}=\frac{0,672}{22,4}=0,3\left(mol\right)\)

Vì \(n_{H_2}>n_{Cl_2}\Leftrightarrow n_{H_2}\) dư, \(n_{Cl_2}\) hết

Theo pt (2) : \(n_{HCl}=n_{AgCl}=0,01\left(mol\right)\) (trong 5g)

\(\Leftrightarrow n_{HCl}=0,01.4=0,04\left(mol\right)\) (trong 20g)

Theo pt (1) : \(n_{Cl_2pư}=\frac{1}{2}n_{HCl}=0,02\left(mol\right)\)

\(H=\frac{0,02}{0,03}.100\%=66,67\%\)

Vậy....

Ta có :

\(A=-7+x-x^2\)

\(=-\left(x^2-x+7\right)\)

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

Với mọi x ta có :

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)

\(\Leftrightarrow-\left[\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\right]\le\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

Vậy...

\(P=\frac{x}{x+1}+\frac{x}{x-1}+\frac{2x^2}{1-x^2}\)

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

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

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

\(=0\)

\(\Leftrightarrow P\) không phụ thuộc vào x \(\left(đpcm\right)\)

Ta có : \(a+b+c=1\Leftrightarrow a+b=1-c\)

Lại có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{1-c}{c}=0\)

\(\Leftrightarrow\frac{1-c}{ab}+\frac{1-c}{c}=0\)

\(\Leftrightarrow\left(1-c\right)\left(\frac{1}{ab}+\frac{1}{c}\right)=0\)

\(\Leftrightarrow\left(1-c\right).\frac{c+ab}{abc}=0\)

\(\Leftrightarrow\left(1-c\right).\frac{1-a-b+ab}{abc}=0\)

\(\Leftrightarrow\frac{\left(1-c\right)\left(1-a\right)\left(1-b\right)}{abc}=0\)

\(\Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)=0\left(đpcm\right)\)

1/ Với mọi x ta có :

\(\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|-4\ge-4\)

\(\Leftrightarrow A\ge-4\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{3}\)

Vậy...

2/ Với mọi x ta có :

\(\left|x-2\right|\ge0\)

\(\Leftrightarrow4\left|x-2\right|\ge0\)

\(\Leftrightarrow-4\left|x-2\right|\le0\)

\(\Leftrightarrow10-4\left|x-2\right|\le10\)

\(\Leftrightarrow B\le10\)

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

Vậy..

3/ Với mọi x ta có :

\(\left|x\right|\ge0\)

\(\Leftrightarrow\left|x\right|-3\ge-3\)

\(\Leftrightarrow\frac{6}{\left|x\right|-3}\le-2\)

\(C\le-2\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy...

Ta có :

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

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^3=a^3\)

\(\Leftrightarrow x^3+3x^2.\frac{1}{x}+3x.\frac{1}{x^2}+\frac{1}{x^3}=a^3\)

\(\Leftrightarrow\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)=a^3\)

\(\Leftrightarrow x^3+\frac{1}{x^3}+3a=a^3\)

\(\Leftrightarrow x^3+\frac{1}{x^3}=a^3-3a\)

Lại có : \(x+\frac{1}{x}=a\)

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

\(\Leftrightarrow x^2+2.x.\frac{1}{x}+\frac{1}{x^2}=a^2\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)

Ta có : \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=\left(a^3-3a\right)\left(a^2-2\right)\)

\(\Leftrightarrow x^5+\frac{1}{x}+x+\frac{1}{x^5}=a^5-2a^3-3a^3+6a\)

\(\Leftrightarrow x^5+\frac{1}{x^5}=a^5-5a^3+5a\)

Vậy.....

Xét

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) \(\left(II\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge9\)

\(\Leftrightarrow1+1+1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}++\frac{c}{a}+\frac{c}{b}\ge9\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)

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

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

+) \(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}.\frac{c}{a}}=2\left(2\right)\)

+) \(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}.\frac{c}{b}}=2\left(3\right)\)

Cộng vế với vế của các BĐT \(\left(1\right),\left(2\right),\left(3\right)\) ta có :

\(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2+2+2\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)\ge6\left(I\right)\)

Vì các BĐT \(\left(I\right);\left(II\right)\) là tương đương nên BĐT (I) luôn đúng \(\Leftrightarrow\) BĐT (2) luôn đúng

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

Vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(đpcm\right)\)

Ta có :

\(M=\frac{1}{x^2+4x+6}\)

\(=\frac{1}{\left(x^2+4x+4\right)+2}\)

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

Với mọi x ta có :

\(\left(x+2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+2\ge2\)

\(\Leftrightarrow\frac{1}{\left(x+2\right)^2+2}\le\frac{1}{2}\)

\(\Leftrightarrow M\le\frac{1}{2}\)

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

Vậy...

Ta có :

\(B=\frac{x^3}{x+1}+\frac{x^2}{x-3}+\frac{1}{x+1}-\frac{9}{x-3}\)

\(=\frac{x^3+1}{x+1}+\frac{x^2-9}{x-3}\)

\(=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}+\frac{\left(x-3\right)\left(x+3\right)}{x-3}\)

\(=x^2-x+1+x+3\)

\(=x^2+4\)

Với mọi x ta có :

\(x^2\ge0\)

\(\Leftrightarrow x^2+4\ge4\)

\(\Leftrightarrow B\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy...

Đề có nhầm gì k bạn ?

Ta có :

\(P=\frac{1}{x-5\sqrt{x}+7}=\frac{1}{\left(\sqrt{x}-\frac{5}{2}\right)^2+\frac{3}{4}}\)

Với mọi x ta có :

\(\left(\sqrt{x}-\frac{5}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{\left(\sqrt{x}-\frac{5}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)

\(\Leftrightarrow P\le\frac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{5}{2}=0\Leftrightarrow\sqrt{x}=\frac{5}{2}\Leftrightarrow x=\frac{25}{4}\)

Vậy....

a/ Áp suất lên đáy bình là :

\(p=d.h=10000.0,12=1200\left(Pa\right)\)

b/ Gọi A và B là 2 điểm nằm trên mặt phân cách của dầu mà nước

Ta có :

\(p_A=p_B\)

\(\Leftrightarrow d_d.h_d=d_n.\left(h_d-h'\right)\)

\(\Leftrightarrow8000.0,08=10000.\left(0,08-h'\right)\)

\(\Leftrightarrow h'=0,016\left(m\right)\)

Vậy....

Ta có :

\(Q=x^2-4x+7\)

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

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

Với mọi x ta có :

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

\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)

\(\Leftrightarrow Q\ge3\)

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

Vậy...