TT
H24
21 tháng 12 2021 lúc 21:33

\(a,2x\left(x+3\right)-x^2-9=0\\ \Leftrightarrow2x^2+6x-x^2-9=0\\ \Leftrightarrow x^2+6x-9=0\\ \Leftrightarrow\left(x^2+6x+9\right)-18=0\\ \Leftrightarrow\left(x+3\right)^2-\left(3\sqrt{2}\right)^2=0\\ \Leftrightarrow\left(x+3+3\sqrt{2}\right)\left(x+3-3\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3-3\sqrt{2}\\x=-3+3\sqrt{2}\end{matrix}\right.\)

\(b,2x^2-x=0\\ \Leftrightarrow x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,x^2+x-30=0\\ \Leftrightarrow\left(x^2+6x\right)-\left(5x+30\right)=0\\\Leftrightarrow x\left(x+6\right)-5\left(x+6\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)

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NT
21 tháng 12 2021 lúc 21:31

a: =>x+3=0

hay x=-3

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