Question

Giải Phương Trình

Answer 1

a: \(6\left(x-1\right)-5=3x\)

=>\(6x-6-5-3x=0\)

=>3x-11=0

=>3x=11

=>\(x=\dfrac{11}{3}\)

b: \(\left(2x+5\right)^2-\left(x-2\right)^2=0\)

=>\(\left(2x+5+x-2\right)\left(2x+5-x+2\right)=0\)

=>(3x+3)(x+7)=0

=>3(x+1)(x+7)=0

=>(x+1)(x+7)=0

=>\(\left[{}\begin{matrix}x+1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-7\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

=>\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{-5x+2}{\left(x-2\right)\cdot\left(x+2\right)}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-5x+2}{\left(x-2\right)\left(x+2\right)}\)

=>\(x^2-3x+2-x^2-2x=-5x+2\)

=>-5x+2=-5x+2

=>0x=0(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}