Bài 1
b) x2+6x-7=0
⇒x2+7x-x-7=0
⇒(x2-x)+(7x-7)=0
\(\Rightarrow\)x(x-1)+7(x-1)=0
⇒(x-1)(x+7)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy S={1;-7}
Bài 1:
a: Ta có: \(\left|5-x\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=2x-3\left(x\le5\right)\\x-5=2x-3\left(x>5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\-x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
b: Ta có: \(x^2+6x-7=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)
d: Ta có: \(\dfrac{7\left(x-2\right)}{6}-2>\dfrac{2\left(x+1\right)}{3}\)
\(\Leftrightarrow7x-14-12>4x+4\)
\(\Leftrightarrow7x-26>4x+4\)
\(\Leftrightarrow3x>30\)
hay x>10
Bài 2:
a: Ta có: \(A=\left(\dfrac{3x^2}{x^2-4}-\dfrac{3}{x+2}+\dfrac{3}{2-x}\right):\dfrac{x+3}{x+2}\)
\(=\dfrac{3x^2-3\left(x-2\right)-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x+3}\)
\(=\dfrac{3x^2-3x+6-3x-6}{x-2}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{3x^2-6x}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{3x}{x+3}\)
b: Ta có: |x-2|=4
\(\Leftrightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=6 vào A, ta được:
\(A=\dfrac{3\cdot6}{6+3}=\dfrac{18}{9}=2\)
c: Để A nguyên thì \(3x⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;3;-3;9;-9\right\}\)
\(\Leftrightarrow x\in\left\{-2;-4;0;-6;6;-12\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-4;0;-6;6;-12\right\}\)
Bài 3:
a: Ta có: \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right)\cdot\dfrac{x^4-2x^2+1}{2}\)
\(=\left(\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\right)\cdot\dfrac{\left(x^2-1\right)^2}{2}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x+1\right)^2\cdot\left(x-1\right)}\cdot\dfrac{\left(x+1\right)^2\cdot\left(x-1\right)^2}{2}\)
\(=\dfrac{-2x\cdot\left(x-1\right)}{2}\)
\(=-x^2+x\)
b: Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Thay x=2 vào A, ta được:
\(A=-4+2=-2\)
