Câu hỏi của KIEU TRANG DOAN THI

Question

1. tìm x , biết

a, $\left(x+2\right)^2-9=0$

b, $\left(x+2\right)^{^2}-x^2+4=0$

Mysterious Person · Accepted Answer

a) $\left(x+2\right)^2-9=0$ $\Leftrightarrow$ $\left(x+2\right)^2=9$

$\Leftrightarrow$ $\left\{{}\begin{matrix}x+2=3\x+2=-3\end{matrix}\right.$ $\Leftrightarrow$ $\left\{{}\begin{matrix}x=1\x=-5\end{matrix}\right.$ vậy $x=1;x=-5$

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

$\Leftrightarrow$ $4x+8=0$ $\Leftrightarrow$ $4x=-8\Leftrightarrow x=-2$ vậy $x=-2$Câu trả lời: a) $\left(x+2\right)^2-9=0$ $\Leftrightarrow$ $\left(x+2\right)^2=9$

$\Leftrightarrow$ $\left\{{}\begin{matrix}x+2=3\x+2=-3\end{matrix}\right.$ $\Leftrightarrow$ $\left\{{}\begin{matrix}x=1\x=-5\end{matrix}\right.$ vậy $x=1;x=-5$

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

$\Leftrightarrow$ $4x+8=0$ $\Leftrightarrow$ $4x=-8\Leftrightarrow x=-2$ vậy $x=-2$

T.Thùy Ninh · Answer

$a,\left(x+2\right)^2-9=0$

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

$\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\x=-5\end{matrix}\right.$$b,\left(x-2\right)^2-x^2+4=0$

$\Leftrightarrow x^2-4x+4-x^2+4=0$

$\Leftrightarrow-4x+8=0\Leftrightarrow-4x=-8\Rightarrow x=2$Câu trả lời: $a,\left(x+2\right)^2-9=0$

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

$\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\x=-5\end{matrix}\right.$$b,\left(x-2\right)^2-x^2+4=0$

$\Leftrightarrow x^2-4x+4-x^2+4=0$

$\Leftrightarrow-4x+8=0\Leftrightarrow-4x=-8\Rightarrow x=2$

Bài 5: Những hằng đẳng thức đáng nhớ (Tiếp)

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