Question

Решите уравнения: (x-2)^2+(x-3)^2=2x^2

5+2x=11-7(x-2)

8+3(2x-6)=9

(-5x-3)(2x-1)=0

(x-6)(4x-6)=0

Answer 1

1)

$\displaystyle \tt (x-2)^2+(x-3)^2=2x^2\\\displaystyle \tt x^2-4x+4+x^2-6x+9=2x^2\\\displaystyle \tt 2x^2-10x+13=2x^2\\\displaystyle \tt -10x+13=0\\\displaystyle \tt -10x=-13\\\displaystyle \tt x=-13\div(-10)\\\displaystyle \tt \bold{x=1,3}$

2)

$\displaystyle \tt 5+2x=11-7(x-2)\\\displaystyle \tt 5+2x=11-7x+14\\\displaystyle \tt 5+2x=25-7x\\\displaystyle \tt 2x+7x=25-5\\\displaystyle \tt 9x=20\\\displaystyle \tt x=20\div9\\\\ \displaystyle \tt \bold{x=2\frac{2}{9}}$

3)

$\displaystyle \tt 8+3(2x-6)=9\\\displaystyle \tt 8+6x-18=9\\\displaystyle \tt 6x-10=9\\\displaystyle \tt 6x=9+10\\\displaystyle \tt 6x=19\\\displaystyle \tt x=19\cdot6\\\\ \displaystyle \tt \bold{x=3\frac{1}{6}}$

4)

$\displaystyle \tt (-5x-3)(2x-1)=0\\\\ \displaystyle \tt -5x-3=0\\\displaystyle \tt -5x=3\\\displaystyle \tt x=3\div(-5)\\\displaystyle \tt \bold{x_1=-0,6}\\\\\displaystyle \tt 2x-1=0\\\displaystyle \tt 2x=1\\\displaystyle \tt x=1\div2\\\displaystyle \tt \bold{x_2=0,5}$

5)

$\displaystyle \tt (x-6)(4x-6)=0\\\\ \displaystyle \tt x-6=0\\\displaystyle \tt \bold{x_1=6}\\\\ \displaystyle \tt 4x-6=0\\\displaystyle \tt 4x=6\\\displaystyle \tt x=6\div4\\\displaystyle \tt \bold{x_2=1,5}$