Ai giúp mình câu này với,mình tick cho
a)\(\left(-\frac{1}{4}\right)^2x-\frac{\sqrt{9}}{8}=\sqrt{\frac{1}{16}}\) b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-150\%=\left(-\frac{1}{2}\right)^2\)
c)\(2x:\frac{7}{4}=35\%:\left(-1\frac{2}{3}\right)^2\) d)\(\left(\frac{1}{2}x^2-18\right)\left(\frac{2}{3}x-2x+\frac{1}{3}\right)=0\)
a) \(\left(-\frac{1}{4}\right)^2x-\frac{\sqrt{9}}{8}=\sqrt{\frac{1}{16}}\)
\(\frac{1}{16}x-\frac{3}{8}=\frac{1}{4}\)
\(\frac{1}{16}x=\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\)
\(x=\frac{5}{8}:\frac{1}{16}=\frac{5}{8}\cdot\frac{16}{1}=10\)
b) \(2\text{|}\frac{1}{2}x-\frac{1}{3}\text{|}-150\%=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(2\text{|}\frac{1}{2}x-\frac{1}{3}\text{|}=\frac{1}{4}+150\%=\frac{1}{4}+\frac{3}{2}=\frac{1}{4}+\frac{6}{4}=\frac{7}{4}\)
\(\text{|}\frac{1}{2}x-\frac{1}{3}\text{|}=\frac{7}{4}:2=\frac{7}{4}\cdot\frac{1}{2}=\frac{7}{8}\)
\(\text{ }\text{ }\frac{1}{2}x-\frac{1}{3}\text{ }=\text{±}\frac{7}{8}\)
TH1: \(\text{ }\text{ }\frac{1}{2}x-\frac{1}{3}\text{ }=\frac{7}{8}\)
\(\text{ }\text{ }\frac{1}{2}x\text{ }=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\)
\(\text{ }\text{ }x\text{ }=\frac{29}{24}:\frac{1}{2}=\frac{29}{24}.2=\frac{29}{12}\)
TH2: \(\text{ }\text{ }\frac{1}{2}x-\frac{1}{3}\text{ }=-\frac{7}{8}\)
\(\text{ }\text{ }\frac{1}{2}x\text{ }=\left(-\frac{7}{8}\right)+\frac{1}{3}=\frac{-13}{24}\)
\(\text{ }\text{ }x\text{ }=\frac{-13}{24}:\frac{1}{2}=\frac{-13}{24}.2=\frac{-13}{12}\)
Vậy \(x\in\left\{\frac{-13}{12};\frac{29}{12}\right\}\)
Tick nha, mình làm 2 bài còn lại cho
nhiều quá bạn đăng từng câu đi mới trả lời được chứ mới nhìn vào là không muốn làm rồi