Câu hỏi của Edogawa Conan

Question

1.Tìm X thuộc Z biết:

a.(-4).$\left(x-2\right)^2=-100$

b.3.(2.X+8)-(5.X+2)=0

c. 5.(7-3.X)+7.(2+2.X)=0

2.CM các đẳng thức sau:

a) x.(y+z)-y.(x-z)=(x+y).z

b) x.(y-z)-x.(y+a)= -x.(z+a)

Nguyễn Thanh Hằng · Accepted Answer

a/ $\left(-4\right)\left(x-2\right)^2=-100$

$\Leftrightarrow\left(x-2\right)^2=\left(-100\right):\left(-4\right)$

$\Leftrightarrow\left(x-2\right)^2=25$

$\Leftrightarrow\left[{}\begin{matrix}x-2=5\x-2=-5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=7\x=-3\end{matrix}\right.$

Vậy ...

b/ $3\left(2x+8\right)-\left(5x+2\right)=0$

$\Leftrightarrow6x+24=5x+2$

$\Leftrightarrow6x-5x=2-24$

$\Leftrightarrow x=-22$

Vậy ....

c/ $5\left(7-3x\right)+7\left(2+2x\right)=0$

$\Leftrightarrow5\left(7-3x\right)=-7\left(2+2x\right)$

$\Leftrightarrow35-15x=-14+\left(-14\right)x$

$\Leftrightarrow35+14=\left(-14\right)x+15x$

$\Leftrightarrow x=49$

Vậy ....Câu trả lời: a/ $\left(-4\right)\left(x-2\right)^2=-100$

$\Leftrightarrow\left(x-2\right)^2=\left(-100\right):\left(-4\right)$

$\Leftrightarrow\left(x-2\right)^2=25$

$\Leftrightarrow\left[{}\begin{matrix}x-2=5\x-2=-5\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=7\x=-3\end{matrix}\right.$

Vậy ...

b/ $3\left(2x+8\right)-\left(5x+2\right)=0$

$\Leftrightarrow6x+24=5x+2$

$\Leftrightarrow6x-5x=2-24$

$\Leftrightarrow x=-22$

Vậy ....

c/ $5\left(7-3x\right)+7\left(2+2x\right)=0$

$\Leftrightarrow5\left(7-3x\right)=-7\left(2+2x\right)$

$\Leftrightarrow35-15x=-14+\left(-14\right)x$

$\Leftrightarrow35+14=\left(-14\right)x+15x$

$\Leftrightarrow x=49$

Vậy ....