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1) \(x^2-3x+2+\left|x-1\right|=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)+\left|x-1\right|=0\) (2)
Xét : \(x< 1\) thì pt (2) trở thành :
\(\left(x-1\right)\left(x-2\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\) ( loại do x < 1 )
Xét \(x\ge1\) pt (2) thở thành :
\(\left(x-1\right)\left(x-2\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=1\) ( thỏa mãn )
Vậy : \(x=1\) thỏa mãn pt đã cho.
a/ - Với \(x\ge1\):
\(\Leftrightarrow x^2-3x+2+x-1=0\)
\(\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)
- Với \(x< 1\)
\(\Leftrightarrow x^2-3x+2+1-x=0\)
\(\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=3\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=1\)
b/ ĐKXĐ: ...
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}\right)+16+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow\left(x+4\right)^2=16\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
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