Question

Tìm x , biết

a) ( x + 6 ) \(^2\) = x + 6

b) ( x + 8 )\(^2\) = 121

c) 2( x + 3 ) - x\(^2\) - 3x = 0

Answer 1

a) \(\left(x+6\right)^2=x+6\)

\(\left(x+6\right)^2-\left(x+6\right)=0\)

\(\left(x+6\right)\left(x+6-1\right)=0\)

\(\left(x+6\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=-5\end{matrix}\right.\)

b) \(\left(x+8\right)^2=121=\left(\pm11\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)

c) \(2\left(x+3\right)-x^2-3x=0\)

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

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

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

Answer 2

a)\(\left(x+6\right)^2=x+6\)

\(\Leftrightarrow x^2+12x+6-x-6=0\)

\(\Leftrightarrow x^2-11x=0\)

\(\Leftrightarrow x\left(x-11\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-11=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)

b)\(\left(x+8\right)^2=121\)

\(\Leftrightarrow\left(x+8\right)^2=11^2\)

\(\Leftrightarrow x+8=11\)

\(\Rightarrow x=3\)

c)\(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Answer 3

a/ \(\left(x+6\right)^2=x+6\)

\(\Leftrightarrow\left(x+6\right)^2-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+6-1\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-5\end{matrix}\right.\)

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b/ \(\left(x+8\right)^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=11\\x+8=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

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c/ \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2x+6-x^2-3x=0\)

\(\Leftrightarrow6-x^2-x=0\)

\(\Leftrightarrow-\left(x^2+x-6\right)=0\)

\(\Leftrightarrow-\left(x^2-2x+3x-6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy..