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Trả lời:
a, \(P=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{1}{\sqrt{x}+1}-\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{2}{x-1}\right)\) \(\left(ĐK:x\ge0;x\ne1\right)\)
\(=\left[\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\right)\)
\(=\left[\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{2}{x-1}\right]\)
\(=\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{x-1}:\frac{\sqrt{x}-1-\sqrt{x}\left(\sqrt{x}+1\right)+2}{x-1}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}:\frac{\sqrt{x}-1-x-\sqrt{x}+2}{x-1}\)
\(=\frac{4\sqrt{x}}{x-1}:\frac{1-x}{x-1}=\frac{4\sqrt{x}}{x-1}\cdot\frac{x-1}{1-x}=\frac{4\sqrt{x}}{1-x}\)
theo mình thì biến đổi cái phương trình đầu rồi dùng bđt để a=b
sau đó thay vào cái thứ 2 là được
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
1) a) \(\hept{\begin{cases}2x-y=5\\x+y=4\end{cases}}\)<=> \(\hept{\begin{cases}3x=9\\x+y=4\end{cases}}\)<=>\(\hept{\begin{cases}x=3\\3+y=4\end{cases}}\)<=> \(\hept{\begin{cases}x=3\\y=1\end{cases}}\)
\(16x^5-8x^3+x=0\)(1) <=> \(x\left(16x^4-8x^2+1\right)=0\)
<=> \(x_1=0\)hoac \(16x^4-8x^2+1=0\)
\(16x^4-8x^2+1=0\)
Dat \(x^2=t\left(t\ge0\right)\)phuong trinh tro thanh
\(16x^2-8x+1=0\)
\(\left(a=16;b'=\frac{b}{2}=-\frac{8}{2}=-4:c=1\right)\)
\(\Delta'=b'^2-ac=\left(-4\right)^2-16\cdot1=16-16=0\)
Phuong trinh co nghiem kep t1 =t2=\(-\frac{b'}{a}=-\frac{-4}{1}=4\)(thoa)
Voi t=4 ta duoc
\(x^2=4\)<=> \(x_2=2,x_3=-2\)
Vay nghiem cua phuong trinh (1) la \(x_1=0,x_2=2,x_3=-2\)
\(A=\left(\frac{\sqrt{X}}{\sqrt{X}+1}+\frac{\sqrt{X}+1}{1-\sqrt{X}}+\frac{4\sqrt{X}+1}{X-1}\right)\left(\frac{X\sqrt{X}}{\sqrt{X}+1}-\sqrt{X}\right)\)
\(=\left(\frac{\sqrt{X}-\sqrt{X}-1+4\sqrt{X}+1}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\right)\left(X-\sqrt{X}\right)\)
\(=\frac{4\sqrt{X}}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}.\sqrt{X}\left(\sqrt{X}-1\right)\)
\(A=\frac{4X}{\sqrt{X}+1}\)
B) dễ rồi làm tiếp ik chỉ cần biến về \(\left(a+b\right)^2+hs\le hs\) là được
\(=\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
=b-a