Question

[phung tu uyen]

tim x biet

cau hoi:

a) #Hỏi cộng đồng OLM #Toán lớp 7

Answer 1

[Nguyễn Trần Diệu Linh]

a) |2,5 - x| = 1,3 |2,5 - x| = 1,3

=> |2,5 - x| = |2,5 - x| = 1,3 : 1,3

=> |2,5 - x| = |2,5 - x| = 1

=> \(\left[{}\begin{matrix}2,5-x=1\\2,5-x=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,5\\x=3,5\end{matrix}\right.\)

Vậy x = 1,5 hoặc x = 3,5

b) 1,6 - |x - 0,2| = 0

=> |x - 0,2| = 1,6

=> \(\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

Vậy x = 1,8 hoặc x = -1,4

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

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

=> \(x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

d) (x - 2)² = 1

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

Vậy x = 3 hoặc x = 1

e) (2x - 1)³ = -8

=> (2x - 1)³ = (-2)³

=> 2x - 1 = -2

=> 2x = -1

=> x = \(\dfrac{-1}{2}\)

Vậy \(x=\dfrac{-1}{2}\)

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

=> \(\left(x+\dfrac{1}{2}\right)^2\) = \(\left(\pm\dfrac{1}{4}\right)^2\)

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

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

Answer 2

[Hebico may mắn]

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

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

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

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

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

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

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

Vậy \(x\in\left\{1;3\right\}\)

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

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\\ \Rightarrow2x-1=-2\\ \Rightarrow2x=-2+1\)

\(\Rightarrow2x=-1\)

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

Vậy \(x=\dfrac{-1}{2}\)

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

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

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

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

Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{3}{4}\right\}\)