bài 8:
a: |x-1|=2
=>\(\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Thay x=3 vào A, ta được:
\(A=\dfrac{3^2-2\cdot3}{3+1}=\dfrac{3}{4}\)
b: \(P=A\cdot B\)
\(=\dfrac{x\left(x-2\right)}{x+1}\left(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}-\dfrac{16}{4-x^2}\right)\)
\(=\dfrac{x\left(x-2\right)}{x+1}\cdot\left(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}+\dfrac{16}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\dfrac{x\left(x-2\right)}{x+1}\cdot\dfrac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x}{x+1}\cdot\dfrac{x^2+4x+4-x^2+4x-4+16}{x+2}\)
\(=\dfrac{x\cdot8\cdot\left(x+2\right)}{\left(x+2\right)\left(x+1\right)}=\dfrac{8x}{x+1}\)
c: P<8
=>P-8<0
=>\(\dfrac{8x}{x+1}-8< 0\)
=>\(\dfrac{8x-8x-8}{x+1}< 0\)
=>\(-\dfrac{8}{x+1}< 0\)
=>x+1>0
=>x>-1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-1\\x< >2\end{matrix}\right.\)
Bài 7:
a: \(4x^2=1\)
=>\(x^2=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(loại\right)\\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Thay x=1/2 vào B, ta được:
\(B=\dfrac{\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}}{2\cdot\dfrac{1}{2}+1}=\dfrac{\dfrac{1}{4}-\dfrac{1}{2}}{2}=\dfrac{-1}{4}:2=\dfrac{-1}{8}\)
b: \(M=A\cdot B\)
\(=\left(\dfrac{1}{x-1}-\dfrac{x}{1-x^2}\right)\cdot\dfrac{x^2-x}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+1}\)
\(=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}=\dfrac{x}{x+1}\)
c: M<1
=>M-1<0
=>\(\dfrac{x-x-1}{x+1}< 0\)
=>\(-\dfrac{1}{x+1}< 0\)
=>x+1>0
=>x>-1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-1\\x\notin\left\{1;-\dfrac{1}{2}\right\}\end{matrix}\right.\)
