-12 - 2x=-8

=> -2x= 4

=> x=-2

Vậy x=-2

\(-12-2x=-8\)

\(2x=-12-(-8)\)

\(2x=-4\)

\(x=-4\div2\)

\(x=-2\)

k cho me nha

Ta có: \(a^2+b^2+c^2=d^2+e^2+g^2\Leftrightarrow a^2+b^2+c^2+d^2+e^2+g^2=2\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2+g^2⋮2\left(1\right)\)

Lại có \(a^2-a=a\left(a-1\right)⋮2\)

Tương tự \(b^2-b,c^2-c,d^2-d,e^2-e,g^2-g⋮2\)

\(\Leftrightarrow\left(a^2+b^2+c^2+d^2+e^2+g^2\right)-\left(a+b+c+d+e+g\right)⋮2\left(2\right)\)

Từ (1) và (2) \(\Leftrightarrow a+b+c+d+e+g⋮2\)

đáp án

ADDBD

Hai thừa số đó là hai số đối nhau.

mÌnh làm sai rồi.

=\(\frac{7}{9}\)

làm thế nào

\(a,234-\left(x-56\right)=789\)

\(\Leftrightarrow x-56=234-789\)

\(\Leftrightarrow x-56=-555\)

\(\Leftrightarrow x=\left(-555\right)+56=-499\)

Vậy x = -499

b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)

\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)

\(\Leftrightarrow4x+12=-5x+75\)

\(\Leftrightarrow4x+12-\left(-5x\right)=75\)

\(\Leftrightarrow4x-\left(-5x\right)+12=75\)

\(\Leftrightarrow4x+5x=63\)

\(\Leftrightarrow9x=63\)

\(\Leftrightarrow x=7\)

Vậy x = 7

c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)

\(\Leftrightarrow8x-8-7=2x+4+5\)

\(\Leftrightarrow8x-8-7-2x+4=5\)

\(\Leftrightarrow8x-2x-8-7+4=5\)

\(\Leftrightarrow8x-2x=5-4+7+8\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\)

Vậy x = 4

d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)

=> \(5⋮x-1\)

=> \(x-1\inƯ\left(5\right)\)

=> \(x-1\in\left\{\pm1;\pm5\right\}\)

=> \(x\in\left\{2;0;6;-4\right\}\)

Vì (x + 3)(x² - 1)=0

=> \(\orbr{\begin{cases}x+3=0\\x^2-1=0\end{cases}}=>\orbr{\begin{cases}x=-3\\x^2=1\end{cases}}=>\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

\(\left(x+3\right)\left(x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2+1=0\end{cases}}\)

Mà \(x^2+1>0\Rightarrow x+3=0\)

\(\Rightarrow x=3\)

Vậy x=3