mẹ có số tiền là:
     5000+10000+20000=35000(đồng)
                   đ/s:35000 đồng k mình nhaaaaaaaaaaaa

10.000 + 30.000 + 80.000 = 120.000

\(\dfrac{5}{12}:\dfrac{3}{4}=\dfrac{5}{12}\times\dfrac{4}{3}=\dfrac{5}{9}\)

5/12:3/4 = 5/12 x 4/3 =10/33

Câu 4 : Gọi số chính phương cần tìm là k^2 ( k thuộc N* ) 

\(n^2-4n+7=k^2\)

\(\Leftrightarrow\left(n-2\right)^2+3-k^2=0\)

\(\Leftrightarrow\left(n-2-k\right)\left(n-2+k\right)=-3\)

\(\Rightarrow n-2-k;n-2+k\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

bạn tự xét từng th này nhé 

`Answer:`

a. \(\left(2x-\frac{1}{3}\right).4=256\)

\(\Leftrightarrow2x-\frac{1}{3}=256:4\)

\(\Leftrightarrow2x-\frac{1}{3}=64\)

\(\Leftrightarrow2x=64+\frac{1}{3}\)

\(\Leftrightarrow2x=\frac{193}{3}\)

\(\Leftrightarrow x=\frac{193}{6}\)

b. \(\left|3x+\frac{1}{2}-\frac{1}{5}\right|=4-\frac{1}{7}\)

\(\Leftrightarrow\left|3x+\frac{1}{2}-\frac{1}{5}\right|=\frac{27}{7}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+\frac{1}{2}-\frac{1}{5}=\frac{27}{7}\\3x+\frac{1}{2}-\frac{1}{5}=-\frac{27}{7}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+\frac{1}{2}=\frac{27}{7}+\frac{1}{5}\\3x+\frac{1}{2}=-\frac{27}{7}+\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+\frac{1}{2}=\frac{142}{35}\\3x+\frac{1}{2}=-\frac{128}{35}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=\frac{142}{35}-\frac{1}{2}\\3x=-\frac{128}{35}-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=\frac{249}{70}\\3x=-\frac{291}{70}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{83}{70}\\x=-\frac{97}{70}\end{cases}}\)

=7/6.9/14

=3/4

chúc bạn học tốt

\(\left(\dfrac{5}{2}-\dfrac{4}{3}\right).\dfrac{9}{14}=\left(\dfrac{15}{6}-\dfrac{8}{6}\right).\dfrac{9}{14}=\dfrac{7}{6}.\dfrac{9}{14}=\dfrac{3}{4}\)

100+100+100000090=100000290 

=100.000.290

\(\dfrac{36}{73}< \dfrac{24}{x}< \dfrac{12}{23}\)

\(\dfrac{72}{146}< \dfrac{72}{3x}< \dfrac{72}{138}\)

\(146>3x>138\)

\(48,6>x>46\)

Vậy \(x=47\) hoặc \(x=48\)