\(\left(\dfrac{123}{41}-6\dfrac{2}{7}+2024^2\right).\left(\dfrac{4}{3}.\dfrac{-1}{6}+\dfrac{5}{6}.\dfrac{-4}{3}+1\dfrac{1}{3}\right)-5\)
2,tìm x, biết:
\(a,\dfrac{5}{12}.x+\dfrac{-7}{4}=\dfrac{2}{3}\) \(b,0.8.\left(x-1\dfrac{4}{5}\right)=\dfrac{3}{10}+20\%\) \(c,\left(x-\dfrac{3}{4}\right).\left(2x+0,8\right)=0\)
Bài 1:
$=(\frac{123}{41}-6\frac{2}{7}+2024^2)\left[\frac{4}{3}(\frac{-1}{6}+\frac{-5}{6})+\frac{4}{3}\right]-5$
$=(\frac{123}{41}-6\frac{2}{7}+2024^2)(\frac{-4}{3}+\frac{4}{3})-5$
$=(\frac{123}{41}-6\frac{2}{7}+2024^2).0-5=0-5=-5$
Lời giải:
a.
$\frac{5}{12}x=\frac{2}{3}-\frac{-7}{4}=\frac{29}{12}$
$x=\frac{29}{12}: \frac{5}{12}=\frac{29}{5}$
b.
$0,8(x-1\frac{4}{5})=\frac{3}{10}+20\text{%}=0,5$
$x-\frac{9}{5}=0,5:0,8=\frac{5}{8}$
$x=\frac{5}{8}+\frac{9}{5}$
$x=\frac{97}{40}$
c.
$(x-\frac{3}{4})(2x+0,8)=0$
$\Rightarrow x-\frac{3}{4}=0$ hoặc $2x+0,8=0$
$\Rightarrow x=\frac{3}{4}$ hoặc $2x=-0,8$
$\Rightarrow x=\frac{3}{4}$ hoặc $x=-0,4$