Câu hỏi của Rachel Kim

Question

Tìm x:

a) 2x(x - 3) + 5(x - 3) = 0

b) (x2 - 4) - (x - 2)(3 - 2x) =0

c) (2x + 5)2 = (x + 2)2

d) x2 - 5x + 6 = 0

e) 2x3 + 6x2 = x2 + 3xcâu

f) (x2 - 1)(x + 2) - (x - 2)(x2 + 2x + 4) - 2x2 = 0

Rachel Kim · Accepted Answer

Câu e) là: 2x3 + 6x2 = x2 + 3x nhéCâu trả lời: Câu e) là: 2x3 + 6x2 = x2 + 3x nhé

Cẩm Mịch · Answer

a) $2x\left(x-3\right)+5\left(x-3\right)=0$

$\Rightarrow\left(x-3\right)\left(2x+5\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-3=0\2x+5=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=3\2x=-5\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=3\x=-\dfrac{5}{2}\end{matrix}\right.$

b) $\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0$

$\Rightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0$

$\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0$

$\Rightarrow\left(x-2\right)\left(3x-1\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-2=0\3x-1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2\3x=1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2\x=\dfrac{1}{3}\end{matrix}\right.$

c) $\left(2x+5\right)^2=\left(x+2\right)^2$

$\Rightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0$

$\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0$

$\Rightarrow\left(x+3\right)\left(3x+7\right)=0$

$\Rightarrow\left[{}\begin{matrix}x+3=0\3x+7=0\end{matrix}\right.$

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

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

d) $x^2-5x+6=0$

$\Rightarrow x^2-2x-3x+6=0$

$\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0$

$\Rightarrow\left(x-2\right)\left(x-3\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-2=0\x-3=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2\x=3\end{matrix}\right.$

e) $2x^3+6x^2=x^2+3x$

$\Rightarrow2x^3+6x^2-x^2-3x=0$

$\Rightarrow2x^3+5x^2-3x=0$

$\Rightarrow x\left(2x^2+5x-3\right)=0$

$\Rightarrow2x^2+5x-3=0$

$\Rightarrow2x^2-6x+x-3=0$

$\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0$

$\Rightarrow\left(x-3\right)\left(2x+1\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-3=0\2x+1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=3\x=-\dfrac{1}{2}\end{matrix}\right.$

f) $\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)-2x^2$

$\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0$

$\Rightarrow x^3+2x^2-x+2-x^3+8-2x^2=0$

$\Rightarrow-x+10=0$

$\Rightarrow x=10$Câu trả lời: a) $2x\left(x-3\right)+5\left(x-3\right)=0$