a: \(-\dfrac{1}{2}+\left(\dfrac{4}{5}+x\right)=1\dfrac{1}{2}\)
=>\(x+\dfrac{4}{5}=\dfrac{3}{2}+\dfrac{1}{2}=2\)
=>\(x=2-\dfrac{4}{5}=\dfrac{6}{5}\)
b: ĐKXĐ: x>=0
\(5\sqrt{x}-\sqrt{\dfrac{1}{25}}=0\)
=>\(5\sqrt{x}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
=>\(\sqrt{x}=\dfrac{1}{25}\)
=>\(x=\left(\dfrac{1}{25}\right)^2=\dfrac{1}{625}\)(nhận)
c: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{3}\\x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
d: ĐKXĐ: x>=0
\(\left(\sqrt{x}-4\right)\left(\sqrt{x+1}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}-4=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x+1}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\left(nhận\right)\\x+1=4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=16\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
Lời giải:
a.
$\frac{-1}{2}+(\frac{4}{5}+x)=1\frac{1}{2}$
$\frac{4}{5}+x=1\frac{1}{2}+\frac{1}{2}=2$
$x=2-\frac{4}{5}=\frac{6}{5}$
b.
$5\sqrt{x}-\sqrt{\frac{1}{25}}=0$
$5\sqrt{x}-\frac{1}{5}=0$
$5\sqrt{x}=\frac{1}{5}$
$\sqrt{x}=\frac{1}{5}:5=\frac{1}{25}$
$x=(\frac{1}{25})^2=\frac{1}{625}$
c.
$(x-\frac{1}{3})^2=\frac{1}{9}=(\frac{1}{3})^2=(\frac{-1}{3})^2$
$\Rightarrow x-\frac{1}{3}=\frac{1}{3}$ hoặc $x-\frac{1}{3}=\frac{-1}{3}$
$\Rightarrow x=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ hoặc $x=\frac{-1}{3}+\frac{1}{3}=0$
d.
$(\sqrt{x}-4)(\sqrt{x+1}-2)=0$
$\Rightarrow \sqrt{x}-4=0$ hoặc $\sqrt{x+1}-2=0$
$\Rightarrow \sqrt{x}=4$ hoặc $\sqrt{x+1}=2$
$\Rightarrow x=4^2=16$ hoặc $x+1=2^2=4$
$\Rightarrow x=16$ hoặc $x=4-1=3$
