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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
V...\(S=\left\{-8\right\}\)
^^
bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)
\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)
\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)
\(\Leftrightarrow23x=69\)
\(\Leftrightarrow x=3\)
Vậy nghiệm của pt x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: x khác 0
Đặt \(x+\frac{1}{x}=a\)\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow a^2=x^2+\frac{1}{x^2}+2\cdot x\cdot\frac{1}{x}\Leftrightarrow a^2-2=x^2+\frac{1}{x^2}\)
Có:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)
\(=8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2\)
\(=8a^2+4\left(a^4-4a^2+4\right)-4\left(a^4-2a^2\right)\)
\(=8a^2+4a^4-16a^2+16-4a^4+8a^2=16\)
Thay \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=16\)
vào phương trình, ta có: \(\left(x-4\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=-4\\x-4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}\)Mà điều kiện x khác 0 nên x=8
Vậy phương trình có nghiệm x=8
\(\frac{x+1}{x-2}-\frac{3-x}{x+2}=\frac{2\left(x^2-2\right)}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow\frac{x+1}{x-2}+\frac{x-3}{x+2}=\frac{2\left(x^2-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-3\right)\left(x-2\right)=2\left(x^2-2\right)\)
\(\Leftrightarrow x^2+3x+2+x^2-5x+6=2x^2+4\)
\(\Leftrightarrow2x^2-2x-2x^2=4-8\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)(không thỏa mãn ĐKXĐ)
Vậy phương trình đã cho vô nghiệm.
\(\frac{x+1}{x-2}-\frac{3-x}{x+2}=\frac{2\left(x^2-2\right)}{x^2-4}\)ĐK : \(x\ne\pm2\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)-\left(3-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+3x+2-3x+6+x^2-2x=2x^2-4\)
\(\Leftrightarrow2x^2-2x+8=2x^2-4\Leftrightarrow-2x+12=0\)
\(\Leftrightarrow-2\left(x-6\right)=0\Leftrightarrow x=6\)( tmđk )
Vậy tập nghiệm của phương trình S= { 6 }