\(\left(\frac{9}{2}-\frac{5}{2}\right)-\frac{3}{4}=\frac{4}{2}-\frac{3}{4}=\frac{8}{4}-\frac{3}{4}=\frac{5}{4}\)\(\frac{5}{4}\)

\(\frac{9}{2}-\left(\frac{5}{2}+\frac{3}{4}\right)=\frac{9}{2}-\left(\frac{10}{4}+\frac{3}{4}\right)=\frac{9}{2}-\frac{13}{4}=\frac{18}{4}-\frac{13}{4}=\frac{5}{4}\)

Vậy : \(\left(\frac{9}{2}-\frac{5}{2}\right)-\frac{3}{4}=\frac{9}{2}-\left(\frac{5}{2}+\frac{3}{4}\right)\)

tk mk nha

Mình cần gấp

sorry,tại nó ko hiện thì nó mới bị thế ở chỗ phép tính 1

A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)

\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)

\(=\dfrac{7}{20}-\dfrac{9}{40}\)

\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)

B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)

\(=\dfrac{5}{6}+\dfrac{11}{36}\).

\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)

C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)

\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)

a: =7/20-5/8+2/5

=14/40-25/40+16/40

=5/40=1/8

b: =5/6+5/9-1/4

=30/36+20/36-9/36

=41/36

c: =9/10-2/5+3/10+7/20

=12/10-2/5+7/20

=7/20+6/5-2/5

=7/20+4/5

=7/20+16/20

=23/20

`a, 3/4 + 1/2 xx 7/2`

`= 3/4 + 7/4`

`=10/4`

`=5/2`

`b, 6/15 - 1/3 : 5/3`

`= 6/15 - 1/3 xx 3/5`

`= 6/15 - 3/15`

`= 3/15`

`=1/5`

`c, x-4/9 = 3/7 : 9/4`

`=> x-4/9= 3/7 xx 4/9`

`=> x-4/9= 12/63`

`=> x-4/9=4/21`

`=> x= 4/21 +4/9`

`=>x= 40/63`

`d, 7/9 xx 3/5 -1/2=1/5`

`->` sao lại bằng có `x` ko vậy ạ?

`a,`

`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`

`b,`

`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`

`c,` Tìm x?

`x-4/9=3/7 \div 9/4`

`x-4/9=4/21`

`x=4/21+4/9`

`x=40/63`

`d, 7/9x \times 3/5-1/2=1/5`

`7/9x \times 3/5=1/5+1/2`

`7/9x \times 3/5=7/10`

`7/9x=7/10 \div 3/5`

`7/9x=7/6`

`x=7/6 \div 7/9=3/2`

Là sao!!!!!???????!!!!!???

\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}=\frac{2}{4}=\frac{1}{2}\)

a: \(x:5=\dfrac{15}{14}\\ \Rightarrow x=\dfrac{15}{70}=\dfrac{3}{14}\)

a) \(x:5=\dfrac{4}{7}+\dfrac{1}{2}\)

\(\dfrac{x}{5}=\dfrac{15}{14}\)

\(\Rightarrow x=\dfrac{5.15}{14}=\dfrac{75}{14}\)

b) \(\dfrac{1}{3}+\dfrac{2}{9}\times\dfrac{3}{4}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{2}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(x\div5=\dfrac{4}{7}+\dfrac{1}{2}\)

\(x\div5=\dfrac{15}{14}\)

\(x=\dfrac{15}{14}\times5\)

\(x=\dfrac{75}{14}\)

b)

\(\dfrac{1}{3}+\dfrac{2}{9}\times\dfrac{3}{4}\)

\(=\dfrac{1}{3}+\dfrac{1}{6}\)

\(=\dfrac{1}{2}\)