Giải phương trình sau: a) $(2x + 5)^{2}$ = $(x + 2)^{2}$ b) $2x^{3}$ + $6x^{2}$ = $x^{2}$ + 3x 07/11/2021 Bởi Parker Giải phương trình sau: a) $(2x + 5)^{2}$ = $(x + 2)^{2}$ b) $2x^{3}$ + $6x^{2}$ = $x^{2}$ + 3x
Đáp án: Giải thích các bước giải: a) $(2x+5)^2=(x+2)^2$ $\to (2x+5-x-2).(2x+5+x+2)=0$ $\to (x+3).(3x+7)=0$ $\to \left[ \begin{array}{l}x=-3\\x=-\dfrac{7}{3}\end{array} \right.$ b) $2x^3+6x^2=x^2+3x$ $\to 2x^2.(x+3)=x.(x+3)$ $\to (x+3).(2x^2-x)=0$ $\to (x+2).x.(2x-1)=0$ $\to \left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\\x=0\end{array} \right.$ Bình luận
`( 2 x + 5 ) ^2 = ( x + 2 ) ^2` `⇔4x^2+20x+25=x^2+4x+4` `⇔4x^2+20x+25-x^2-4x-4=0` `⇔3x^2+9x+7x+21=0` `⇔3x(x+3)+7(x+3)=0=(x+3)(3x+7)=0` ⇔\(\left[ \begin{array}{l}x+3=0\\3x+7=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-3\\x=\dfrac{-7}{3}\end{array} \right.\) Vậy `S={-3;frac{-7}{3}}` Bình luận
Đáp án:
Giải thích các bước giải:
a) $(2x+5)^2=(x+2)^2$
$\to (2x+5-x-2).(2x+5+x+2)=0$
$\to (x+3).(3x+7)=0$
$\to \left[ \begin{array}{l}x=-3\\x=-\dfrac{7}{3}\end{array} \right.$
b) $2x^3+6x^2=x^2+3x$
$\to 2x^2.(x+3)=x.(x+3)$
$\to (x+3).(2x^2-x)=0$
$\to (x+2).x.(2x-1)=0$
$\to \left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\\x=0\end{array} \right.$
`( 2 x + 5 ) ^2 = ( x + 2 ) ^2`
`⇔4x^2+20x+25=x^2+4x+4`
`⇔4x^2+20x+25-x^2-4x-4=0`
`⇔3x^2+9x+7x+21=0`
`⇔3x(x+3)+7(x+3)=0=(x+3)(3x+7)=0`
⇔\(\left[ \begin{array}{l}x+3=0\\3x+7=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=\dfrac{-7}{3}\end{array} \right.\)
Vậy `S={-3;frac{-7}{3}}`