a: \(\left(2x+7\right)^2=4\left(x+2\right)^2\)
=>\(\left(2x+7\right)^2=\left(2x+4\right)^2\)
=>\(\left(2x+7\right)^2-\left(2x+4\right)^2=0\)
=>\(4x^2+28x+49-4x^2-16x-16=0\)
=>12x+33=0
=>12x=-33
=>x=-11/4
b: \(\left(x^2-1\right)\left(x+2\right)=\left(x^2-4\right)\left(x+2\right)\)
=>\(\left(x^2-1\right)\left(x+2\right)-\left(x^2-4\right)\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2-1-x^2+4\right)=0\)
=>3(x+2)=0
=>x+2=0
=>x=-2
d: \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
=>\(\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
mà \(x^2+x+6=x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>=\dfrac{23}{4}\forall x\)
nên \(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
e: \(\left(4x-10\right)\left(5x+24\right)=4x\left(5x-1\right)\)
=>\(20x^2+96x-50x-240=20x^2-4x\)
=>46x-240=-4x
=>50x=240
=>x=4,8
e: \(\left(5x+2\right)\left(x-7\right)=x\left(5x+3\right)\)
=>\(5x^2-35x+2x-14=5x^2+3x\)
=>-33x-14=3x
=>-36x=14
=>\(x=\dfrac{14}{-36}=-\dfrac{7}{18}\)
f: \(\left(4x+2\right)\left(x^2+1\right)=2x^2\left(x+1\right)-5\)
=>\(4x^3+4x+2x^2+2=2x^3+2x^2-5\)
=>\(2x^3+4x+7=0\)
=>\(x\simeq-1,09\)
