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a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
Ta có : \(9^{x-1}=\frac{1}{9}\)
=> \(9^{x-1}=9^{-1}\)
=> x - 1 = -1
=> x = 0
ko biết bạn học mũ âm chưa nêu chưa thì mk xin lỗi
=>
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\frac{4}{3}\)
\(\left|x+\frac{1}{2}\right|=\frac{4}{3}+\frac{2}{3}\)
\(\left|x+\frac{1}{2}\right|=2\)
TH1: \(x+\frac{1}{2}=2\)
\(x=2-\frac{1}{2}\)
\(x=\frac{3}{2}\)
TH2: \(x+\frac{1}{2}=-2\)
\(x=-2-\frac{1}{2}\)
\(x=\frac{-5}{2}\)
KL: x =3/2 hoặc x= -5/2
\(\left|x+\frac{1}{2}\right|-\frac{2}{3}=\sqrt{\frac{16}{9}}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|-\frac{2}{3}=\frac{4}{3}\)
\(\Leftrightarrow\left|x+\frac{1}{2}\right|=\frac{8}{3}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{8}{3}\\x+\frac{1}{2}=-\frac{8}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=-\frac{19}{6}\end{cases}}\)
vậy.....
\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)
\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)
\(=\frac{5}{2}-\frac{8}{27}-8\)
\(=\frac{-313}{54}\)