Đề yêu cầu tìm x à em?
b: \(\Leftrightarrow\left(x-1\right)\left(x+7\right)=\left(x-1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3-x-7\right)=0\)
=>(x-1)(x-10)=0
hay \(x\in\left\{1;10\right\}\)
b: \(\Leftrightarrow\left(x-2\right)\left(-x-1\right)-\left(x-2\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-x-1-3x-5\right)=0\)
=>(x-2)(4x+6)=0
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left(x-5\right)\left(-x-6\right)-\left(x-5\right)\left(7x+8\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(-x-6-7x-8\right)=0\)
=>(x-5)(8x+14)=0
=>x=5 hoặc x=-7/4
b: \(\Leftrightarrow\left(x-4\right)\left(2x+5-5+x\right)=0\)
=>x(x-4)=0
=>x=0 hoặc x=4
b: \(\Rightarrow\left(x-2\right)\left(7-3x-4x+3\right)=0\)
=>(x-2)(10-7x)=0
=>x=2 hoặc x=10/7
\(\left(x-1\right)\left(x+7\right)=\left(1-x\right)\left(3-2x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+7+3-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(10-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
