\(b,\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{3}:\dfrac{1}{3}\)
\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{3}.3\)
\(\left(x-\dfrac{3}{5}\right)^2=4\)
\(\left(x-\dfrac{3}{5}\right)^2=2^2\)
\(x-\dfrac{3}{5}=2\) hoặc \(x-\dfrac{3}{5}=-2\)
\(x=2+\dfrac{3}{5}\) \(x=-2+\dfrac{3}{5}\)
\(x=\dfrac{13}{5}\) \(x=\dfrac{-7}{5}\)
Vậy ...
\(d,\left|x-\dfrac{2}{3}\right|-0,75=1\dfrac{1}{4}\)
\(\left|x-\dfrac{2}{3}\right|=1\dfrac{1}{4}+0,75\)
\(\left|x-\dfrac{2}{3}\right|=\dfrac{5}{4}+\dfrac{3}{4}\)
\(\left|x-\dfrac{2}{3}\right|=\dfrac{8}{4}\)
\(\left|x-\dfrac{2}{3}\right|=2\)
\(x-\dfrac{2}{3}=2\) hoặc \(x-\dfrac{2}{3}=-2\)
\(x=2+\dfrac{2}{3}\) \(x=-2+\dfrac{2}{3}\)
\(x=\dfrac{8}{3}\) \(x=\dfrac{-4}{3}\)
Vậy ...
a: \(\left(3x^2+1\right)\left(4x+\dfrac{1}{3}\right)=0\)
mà \(3x^2+1>=1>0\forall x\)
nên \(4x+\dfrac{1}{3}=0\)
=>\(4x=-\dfrac{1}{3}\)
=>\(x=-\dfrac{1}{12}\)
b: \(\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{3}:\dfrac{1}{3}\)
=>\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{3}\cdot3=4\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{5}=2\\x-\dfrac{3}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\dfrac{3}{5}=\dfrac{13}{5}\\x=-2+\dfrac{3}{5}=-\dfrac{7}{5}\end{matrix}\right.\)
c: \(\left(x+2\sqrt{16}\right)\cdot\left|2x+3\right|=0\)
=>\(\left(x+8\right)\cdot\left|2x+3\right|=0\)
=>\(\left[{}\begin{matrix}x+8=0\\2x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-8\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d:
\(\left|x-\dfrac{2}{3}\right|-0,75=1\dfrac{1}{4}\)
=>\(\left|x-\dfrac{2}{3}\right|=1+\dfrac{1}{4}+0,75=2\)
=>\(\left[{}\begin{matrix}x-\dfrac{2}{3}=2\\x-\dfrac{2}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\dfrac{2}{3}=\dfrac{8}{3}\\x=-2+\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\)
