1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
5.
$x^2-(2x-1)(x+3)=3-x(5+x)$
$\Leftrightarrow x^2-(2x^2+5x-3)=3-(5x+x^2)$
$\Leftrightarro -x^2-5x+3=3-5x-x^2$ (luôn đúng)
Vậy pt có nghiệm $x\in\mathbb{R}$
6.
$3(5-2x)-4(x+2)=5x-18$
$\Leftrightarrow 15-6x-4x-8=5x-18$
$\Leftrightarrow 7-10x=5x-18$
$\Leftrightarrow 25=15x$
$\Leftrightarrow x=\frac{5}{3}$
1: Ta có: \(\left(x-2\right)\left(x-5\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x+10=x^2-7x+12\)(vô lý
2: Ta có: \(\left(x-7\right)\left(x+7\right)+x^2-2=2\left(x^2+5\right)\)
\(\Leftrightarrow x^2-49+x^2-2=2x^2+10\)
\(\Leftrightarrow-51=10\)(vô lý
3: Ta có: \(\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2-8\)
\(\Leftrightarrow4x=-18\)
hay \(x=-\dfrac{9}{2}\)