Question

2 tìm x,biết

a,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) b,\(\left|2x-1\right|=23\)

C,\(\sqrt{x=6}\) d,\(\dfrac{x}{4}=\dfrac{12}{18}\)

Answer 1

a) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=\dfrac{-3}{20}\)

vậy \(x=\dfrac{-3}{20}\)

b) \(\left|2x-1\right|=23\)

\(2x-1=\pm23\)

+) \(2x-1=23\Rightarrow2x=24\Rightarrow x=12\)

+) \(2x-1=-23\Rightarrow2x=-22\Rightarrow x=-11\)

vậy \(x\in\left\{-11;12\right\}\)

Answer 2

c) Xin sửa lại đề:

\(\sqrt{x} = 6\)

=> x = 6²

x = 36

d) \(\frac{x}{4}=\frac{12}{18}\)

=> \(x = \frac{4.12}{18}=\frac{8}{3}\)