1,Tìm x ∈Q:
a,|$\frac{-1}{6}$ |+x=|$\frac{-4}{5}$
b,$\frac{4}{9}$-|x|=$\frac{-1}{27}$
c,$x^{2}$ =$\frac{1}{16}$
d,$x^{2}$ =$\frac{9}{4}$
e,1$\frac{2}{5}$ .x+$\frac{3}{7}$ =$\frac{-4}{5}$
1,Tìm x ∈Q: a,|$\frac{-1}{6}$ |+x=|$\frac{-4}{5}$ b,$\frac{4}{9}$-|x|=$\frac{-1}{27}$ c,$x^{2}$ =$\frac{1}{16}$ d,$x^{2}$ =$\frac{9}{4}$ e,1$\fra
By Mackenzie
Đáp án:
$\begin{array}{l}
a)\left| {\dfrac{{ – 1}}{6}} \right| + x = \left| {\dfrac{{ – 4}}{5}} \right|\\
\Rightarrow \dfrac{1}{6} + x = \dfrac{4}{5}\\
\Rightarrow x = \dfrac{4}{5} – \dfrac{1}{6}\\
\Rightarrow x = \dfrac{{24}}{{30}} – \dfrac{5}{{30}}\\
\Rightarrow x = \dfrac{{19}}{{30}}\\
\text{Vậy}\,x = \dfrac{{19}}{{30}}\\
b)\dfrac{4}{9} – \left| x \right| = \dfrac{{ – 1}}{{27}}\\
\Rightarrow \left| x \right| = \dfrac{4}{9} – \left( {\dfrac{{ – 1}}{{27}}} \right)\\
\Rightarrow \left| x \right| = \dfrac{4}{9} + \dfrac{1}{{27}}\\
\Rightarrow \left| x \right| = \dfrac{{12 + 1}}{{27}}\\
\Rightarrow \left| x \right| = \dfrac{{13}}{{27}}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{13}}{{27}}\\
x = \dfrac{{ – 13}}{{27}}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{{13}}{{27}}\,\text{hoặc}\,x = \dfrac{{ – 13}}{{27}}\\
c){x^2} = \dfrac{1}{{16}}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = – \dfrac{1}{4}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{1}{4}\,\text{hoặc}\,x = \dfrac{{ – 1}}{4}\\
d){x^2} = \dfrac{9}{4}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = – \dfrac{3}{2}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{3}{2}\,\text{hoặc}\,x = – \dfrac{3}{2}\\
e)1\dfrac{2}{5}.x + \dfrac{3}{7} = \dfrac{{ – 4}}{5}\\
\Rightarrow \dfrac{7}{5}.x = – \dfrac{4}{5} – \dfrac{3}{7}\\
\Rightarrow \dfrac{7}{5}x = \dfrac{{ – 28 – 15}}{{35}}\\
\Rightarrow \dfrac{7}{5}x = \dfrac{{ – 43}}{{35}}\\
\Rightarrow x = \dfrac{{ – 43}}{{35}}.\dfrac{5}{7}\\
\Rightarrow x = \dfrac{{ – 43}}{{49}}\\
\text{Vậy}\,x = \dfrac{{ – 43}}{{49}}
\end{array}$