Question

[Hải Linh]

\(\left(\dfrac{123}{41}-6\dfrac{2}{7}+2024^2\right).\left(\dfrac{4}{3}.\dfrac{-1}{6}+\dfrac{5}{6}.\dfrac{-4}{3}+1\dfrac{1}{3}\right)-5\)

2,tìm x, biết:

\(a,\dfrac{5}{12}.x+\dfrac{-7}{4}=\dfrac{2}{3}\)        \(b,0.8.\left(x-1\dfrac{4}{5}\right)=\dfrac{3}{10}+20\%\)  \(c,\left(x-\dfrac{3}{4}\right).\left(2x+0,8\right)=0\)

Answer 1

[Akai Haruma]

Bài 1:

$=(\frac{123}{41}-6\frac{2}{7}+2024^2)\left[\frac{4}{3}(\frac{-1}{6}+\frac{-5}{6})+\frac{4}{3}\right]-5$

$=(\frac{123}{41}-6\frac{2}{7}+2024^2)(\frac{-4}{3}+\frac{4}{3})-5$

$=(\frac{123}{41}-6\frac{2}{7}+2024^2).0-5=0-5=-5$

Answer 2

[Akai Haruma]

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 $x=-0,4$