`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

DÀI QUÁ BẠN Ạ,LÀM ƠN "T I C K" CHO MIK CÁI, THANKS BẠN!

bn ơi ko thì bn làm từng phép một cũng đc nhé các bn giải hộ mình mai mình phải nộp rồi

`4/7+4`

`=4/7+4/1`

`=4/7+28/7`

`=32/7`

__

`3+6/11`

`=33/11+6/11`

`=39/11`

__

`3-5/7`

`=3/1-5/7`

`=21/7-5/7`

`=16/7`

__

`21/9-2`

`=21/9-18/9`

`=3/9`

`=1/3`

__

`15/24+2`

`=15/24+48/24`

`=63/24`

`=21/16`

__

`63/45-20/25`

`=63/45-4/5`

`=63/45-36/45`

`=27/45`

`=9/15`

__

`3/4-2/8`

`=3/4-1/4`

`=2/4`

__

`6/7-5/8`

`=48/56-35/56`

`=13/56`

__

`37/45-5/9`

`=37/45-25/45`

`=12/45`

`=4/15`

__

`46/39-11/13`

`=46/39-33/39`

`=13/39`

`=1/2`

__

`5/12+3/4+1/3`

`=5/12+9/12+4/12`

`=14/12+4/12`

`=18/12`

`=3/2`

__

`1/2+3/7+11/14`

`=7/14+6/14+11/14`

`=13/14+11/14`

`=24/14`

`=12/7`

__

`7/10-(1/5+1/4)`

`=7/10-(4/20+5/20)`

`=7/10-9/20`

`=14/20-9/20`

`=5/20`

`=1/4`

__

`15/4-2/3-3/4`

`=(15/4-3/4)-2/3`

`=12/4-2/3`

`=3-2/3`

`=9/3-2/3`

`=7/3`

\(a,1\dfrac{4}{7}.3\dfrac{4}{11}.3\dfrac{11}{15}.5\dfrac{5}{8}\)

\(=\dfrac{11}{7}.\dfrac{27}{11}.\dfrac{56}{15}.\dfrac{45}{8}\)

\(=\dfrac{11.27.56.45}{7.11.15.8}\)

\(=\dfrac{1.3.7.3}{1.1.1.1}\)

\(=63\)

\(b,\dfrac{3}{4}.1\dfrac{1}{2}+\dfrac{3}{4}.\dfrac{1}{2}\)

\(=\dfrac{3}{4}.\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)\)

\(=\dfrac{3}{4}.2\)

\(=\dfrac{3}{2}\)

Bài 3 là hỗn số hả em?

Bài 1: 

a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)

\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)

hay \(x=-\dfrac{1}{3}\)

Vậy: \(x=-\dfrac{1}{3}\)

b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)

hay \(x=\dfrac{50}{9}\)

Vậy: \(x=\dfrac{50}{9}\)

c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)

\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)

hay \(x=\dfrac{22}{15}\)

Vậy: \(x=\dfrac{22}{15}\)

d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)

\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)

hay \(x=\dfrac{15}{19}\)

Vậy:\(x=\dfrac{15}{19}\)