Question

a,6x²=(2x+5)(3x-5)+7

b,(x+2)²+(x+3)²-2(x-1)(x+1)=9

c,(x-5)²=16

Answer 1

a,Ta có: \(6x^2=\left(2x+5\right)\left(3x-5\right)+7\)

\(=6x^2-10x+15x-25+7\)

\(=6x^2+5x-18\)

\(\Rightarrow5x-18=0\Rightarrow x=\dfrac{18}{5}\)

b, Ta có:\(x^2+4x+4+x^2+6x+9-2x^2+2=9\)

\(\Leftrightarrow10x=-6\Leftrightarrow x=-\dfrac{3}{5}\)

c,Ta có: \(\left(x-5\right)^2=16\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)