Question

Tìm x:

(2x²+x)²-4(2x²+x)+3=0

(x+y)²+(1+x)(1+y)=0

(3x-2)(x+1)²(3x+8)=-16

(4x-5)(2x-3)(x-1)=9

Answer 1

\(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-5=9\\2x-3=9\\x-1=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=6\\x=10\end{matrix}\right.\)

Vậy \(x=\left\{3,5;6;10\right\}\)

Answer 2

d: Sửa đề: \(\left(4x-5\right)^2\cdot\left(2x-3\right)\left(x-1\right)=9\)

a: \(\Leftrightarrow\left(2x^2+x\right)^2-3\left(2x^2+x\right)-\left(2x^2+x\right)+3=0\)

\(\Leftrightarrow\left(2x^2+x\right)\left(2x^2+x-3\right)-\left(2x^2+x-3\right)=0\)

\(\Leftrightarrow\left(2x^2+x-3\right)\left(2x^2+x-1\right)=0\)

\(\Leftrightarrow\left(2x^2+3x-2x-3\right)\left(2x^2+2x-x-1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)

hay \(x\in\left\{-\dfrac{3}{2};1;-1;\dfrac{1}{2}\right\}\)