Question

алгебра 7 класс решить уравнения. Халявщиков блокирую.​

Answer 1

$1)0,5x(x+17)-12(x+17)=0\\\\0,5x^{2}+8,5x-12x-204=0\\\\0,5x^{2}-3,5x-204=0|*2\\\\x^{2}-7x-408=0\\\\D=(-7)^{2}-4*(-408)=49+1632=1681=41^{2}\\\\x_{1}=\frac{7-41}{2}=-17\\\\x_{2} =\frac{7+41}{2}=24\\\\\boxed{x=-17; \ 24}$

Второй способ :

$0,5x(x+17)-12(x+17)=0\\\\(x+17)(0,5x-12)=0\\\\\left[\begin{array}{ccc}x+17=0\\0,5x-12=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1}=-17 \\x_{2}=24 \end{array}\right$

$2)(3x-5)(x-20)+(15-2x)(x-20)=0\\\\3x^{2}-60x-5x+100+15x-300-2x^{2}+40x=0\\\\x^{2} -10x-200=0\\\\\frac{D}{4}=(-\frac{10}{2})^{2}-(-200)=25+200=225=15^{2}\\\\x_{1}=5-15=-10\\\\x_{2}=5+15=20\\\\\boxed{x=-10; \ 20}$

Второй способ:

$(3x-5)(x-20)+(15-2x)(x-20)=0\\(x-20)(3x-5+15-2x)=0\\\\(x-20)(x+10)=0\\\\\left[\begin{array}{ccc}x-20=0\\x+10=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x_{1}=20 \\x_{2}=-10 \end{array}\right\\\\\\3)(7x-28)-4x(7x-28)=0\\\\(7x-28)(1-4x)=0\\\\\left[\begin{array}{ccc}7x-8=0\\1-4x=0\end{array}\right$

$\left[\begin{array}{ccc}7x=28\\4x=1\end{array}\right \\\\\\\left[\begin{array}{ccc}x_{1}=4 \\x_{2}=0,25 \end{array}\right\\\\\boxed{x=4; \ 0,25}$