Ôn tập toán 6

Phương Cao Thanh

Tìm x\(\in\)Q,biết

a)\(\left(x-\dfrac{1}{2}\right)^2=0\)

b)\(\left(x-2\right)^2=1\)

c)\(\left(2x-1\right)^3=-8\)

d)\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

Mới vô
8 tháng 8 2017 lúc 10:10

a,

\(\left(x-\dfrac{1}{2}\right)^2=0\\ \Rightarrow x-\dfrac{1}{2}=0\\ \Rightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b,

\(\left(x-2\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy \(x=3\text{ hoặc }x=1\)

c,

\(\left(2x-1\right)^3=-8\\ \Rightarrow2x-1=-2\\ \Rightarrow2x=-1\\ \Rightarrow x=\dfrac{-1}{2}\)

Vậy \(x=\dfrac{-1}{2}\)

d,

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Vậy \(x=\dfrac{-1}{4}\text{ hoặc }x=\dfrac{-3}{4}\)

Lê Gia Bảo
8 tháng 8 2017 lúc 10:16

a) \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)

\(\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\left(x-2\right)^2=1\Rightarrow\left(x-2\right)^2=1^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=-1\\x-2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1+2\\x=1+2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c) \(\left(2x-1\right)^3=-8\Rightarrow\left(2x-1\right)^3=-2^3\)

\(\Rightarrow2x-1=-3\Rightarrow2x=-3+1\)

\(\Rightarrow2x=2\Rightarrow x=1\)

Vậy \(x=1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)


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