giúp mik bài 2 ạ
Bài 2:
a: \(\dfrac{15}{8}-\dfrac{1}{8}:\left(x-1\right)=\sqrt{\dfrac{9}{16}}\)
=>\(\dfrac{1}{8}:\left(x-1\right)=\dfrac{15}{8}-\dfrac{3}{4}=\dfrac{15}{8}-\dfrac{6}{8}=\dfrac{9}{8}\)
=>\(x-1=\dfrac{1}{8}:\dfrac{9}{8}=\dfrac{1}{9}\)
=>\(x=1+\dfrac{1}{9}=\dfrac{10}{9}\)
b: \(\left|2x-\dfrac{1}{5}\right|\left(2x^2+3\right)=0\)
mà \(2x^2+3>=3>0\forall x\)
nên \(\left|2x-\dfrac{1}{5}\right|=0\)
=>\(2x-\dfrac{1}{5}=0\)
=>\(2x=\dfrac{1}{5}\)
=>\(x=\dfrac{1}{10}\)
c: \(\dfrac{1}{4}-\dfrac{1}{2}:\left|x-2\right|=-\dfrac{1}{2}\)
=>\(\dfrac{1}{2}:\left|x-2\right|=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)
=>\(\left|x-2\right|=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{2}{3}\)
=>\(\left[{}\begin{matrix}x-2=\dfrac{2}{3}\\x-2=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\dfrac{2}{3}=\dfrac{8}{3}\\x=2-\dfrac{2}{3}=\dfrac{4}{3}\end{matrix}\right.\)
d:
ĐKXĐ: x>=0
\(\left(1-\dfrac{3}{2}\left|x-3\right|\right)\left(\sqrt{x}+2\right)=0\)
mà \(\sqrt{x}+2>=2\forall x>=0\)
nên \(1-\dfrac{3}{2}\left|x-3\right|=0\)
=>\(\dfrac{3}{2}\left|x-3\right|=1\)
=>\(\left|x-3\right|=1:\dfrac{3}{2}=\dfrac{2}{3}\)
=>\(\left[{}\begin{matrix}x-3=\dfrac{2}{3}\\x-3=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\left(nhận\right)\\x=\dfrac{7}{3}\left(nhận\right)\end{matrix}\right.\)
Bài 1:
\(a,\sqrt{\dfrac{4}{25}}+\left|-\dfrac{4}{5}\right|-\dfrac{6}{5}.\left(\dfrac{-1}{3}\right)^2\)
\(=\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{6}{5}.\dfrac{1}{9}\)
\(=\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{2}{15}\)
\(=\dfrac{6}{5}-\dfrac{2}{15}\)
\(=\dfrac{18}{15}-\dfrac{2}{15}=\dfrac{16}{15}\)
\(b,\left(-\dfrac{1}{2}\right)^2.\sqrt{\dfrac{16}{25}}+\sqrt{9}.\left(-\dfrac{2}{3}\right)^2-1\dfrac{1}{2}\)
\(=\dfrac{1}{4}.\dfrac{4}{5}+3.\dfrac{4}{9}-\dfrac{3}{2}\)
\(=\dfrac{1}{5}+\dfrac{4}{3}-\dfrac{3}{2}\)
\(=\dfrac{6}{30}+\dfrac{40}{30}-\dfrac{45}{30}=\dfrac{1}{30}\)
\(c,\left(\dfrac{-5}{4}\right)^2.\left|-1\dfrac{1}{5}\right|-\sqrt{\dfrac{4}{81}}:\dfrac{16}{9}+\left(2025\right)^0\)
\(=\dfrac{25}{16}.\left|\dfrac{-6}{5}\right|-\dfrac{2}{9}.\dfrac{9}{16}+1\)
\(=\dfrac{25}{16}.\dfrac{6}{5}-\dfrac{1}{8}+1\)
\(=\dfrac{15}{8}-\dfrac{1}{8}+1\)
\(=\dfrac{14}{8}+1\)
\(=\dfrac{11}{4}\)
\(d,\left(\dfrac{2}{3}-0,75\right):\sqrt{\dfrac{1}{16}}+\left(\dfrac{4}{15}:\dfrac{-8}{25}\right).\sqrt{\dfrac{64}{25}}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{4}\right):\dfrac{1}{4}+\left(\dfrac{4}{15}.\dfrac{-25}{8}\right).\dfrac{8}{5}\)
\(=\left(\dfrac{8}{12}-\dfrac{9}{12}\right).4+\dfrac{-5}{6}.\dfrac{8}{5}\)
\(=\dfrac{-1}{12}.4+\dfrac{-4}{3}\)
\(=\dfrac{-1}{3}-\dfrac{4}{3}\)
\(=\dfrac{-5}{3}\)
