Question

Tìm x ϵ Z biết:

a. (x-1)² = 0

b. x.(x-5)=0

c. x²+4x=0

d. (2x+3)²=49

Answer 1

a/ \(\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2=0^2\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

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b/ \(x\left(x-5\right)=0\)

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

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

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c/ \(x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

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d/ \(\left(2x+3\right)^2=49\)

\(\Leftrightarrow\left(2x+3\right)^2=7^2=\left(-7\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=7\\2x+3=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

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Answer 2

a. (x-1)² = 0

=> x-1=0 => x=1

b. x(x-5) = 0

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

c. x² + 4x = 0

x(x+4) = 0

=>\(\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

d. (2x+3)² = 49

(2x+3)² = \(\left(\pm7\right)^2\)

=>\(\left[{}\begin{matrix}2x+3=7\\2x+3=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Answer 3

a,\(\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

b,\(x\left(x-5\right)=0\)

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

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c,\(x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

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d,\(\left(2x+3\right)^2=49\)

\(\Leftrightarrow2x+3=\pm7\)

\(\Leftrightarrow2x=\pm10\)

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

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