\(=\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=-\dfrac{6}{5}\times\dfrac{5}{7}+\dfrac{13}{5}=\dfrac{61}{35}\)

=\(\dfrac{-3}{5}\).\(\dfrac{5}{7}\)-\(\dfrac{3}{5}\).\(\dfrac{5}{7}\)+\(\dfrac{13}{5}\)

=(\(\dfrac{-3}{5}\)-\(\dfrac{3}{5}\)+\(\dfrac{13}{5}\)).\(\dfrac{5}{7}\)
=\(\dfrac{13}{5}\).\(\dfrac{5}{7}\)
=\(\dfrac{13}{1}\).\(\dfrac{1}{7}\)
=\(\dfrac{13}{7}\)

\(\dfrac{\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{7}}{\dfrac{8}{3}-\dfrac{8}{5}+\dfrac{8}{7}}=\dfrac{2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}{8\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\\ =\dfrac{2}{8}=\dfrac{1}{4}\)

2/3 - 2/5 + 2/7

= 70/105 - 42/105 + 20/105

= (70 - 42 + 20) / 105

= 48 / 105

= 16/35

8/3 - 8/5 + 8/7

= 280/105 - 168 / 105 + 120/105

= ( 280 - 168 + 120 ) / 105

= 232/105

2/3 - 2/5 + 2/7

= 70/105 - 42/105 + 30/105

= (70 - 42 + 30) / 105

= 58 / 105

1, \(\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=\left(\dfrac{-3}{5}-\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)

\(=\dfrac{-6}{5}:\dfrac{7}{5}+\dfrac{13}{5}\)

\(=\dfrac{-6}{7}+\dfrac{13}{5}\\ =\dfrac{61}{35}\)

2, \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\left(-\dfrac{1}{9}\right)\)

\(=0,75-\dfrac{37}{12}+9\cdot\left(-\dfrac{1}{9}\right)\)

\(=-\dfrac{7}{3}+\left(-1\right)\\ -\dfrac{10}{3}\)

cậu lấy máy tính,tính đi ạ

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}=\dfrac{5}{12}.\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)=\dfrac{5}{12}.\left(-1\right)=-\dfrac{5}{12}\)

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}.\left(-\dfrac{9}{2}\right)=-\dfrac{3}{2}\)

a: =-1/3+1/3=0

b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)

c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)

d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)

a: =-1/3+1/3=0

b: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167

d: 

12+3/7-11+3/7 tính bằng cách thuận tiện nhé

a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)

\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)

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

\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)

b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)

\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)

\(=-10\cdot\dfrac{1}{5}=-2\)

c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)

\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)

\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)

d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)

\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)

\(=\dfrac{16}{16}-\dfrac{25}{25}\)

\(=1-1=0\)

e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)

\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)

\(=\dfrac{5}{6}+\dfrac{4}{3}\)

\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(\Rightarrow A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(\Rightarrow A=\left(7-6-5\right)-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}+\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\)

\(\Rightarrow A=-4-\dfrac{1}{4}+0\)

\(\Rightarrow A=-\dfrac{17}{4}\)

Cách 2 tính trong dấu ngoặc đơn của A, bạn tự tính nhé.

A = -17/4

A = -17/4

c) \(\dfrac{11}{10}-\dfrac{-7}{24}=\dfrac{11}{10}+\dfrac{7}{24}=\dfrac{167}{120}\)

e) \(\dfrac{-8}{3}\cdot\dfrac{15}{7}=\dfrac{-120}{21}=\dfrac{-40}{7}\)

f) \(\dfrac{-2}{5}\cdot4\dfrac{1}{2}=\dfrac{-2}{5}\cdot\dfrac{9}{2}=-\dfrac{9}{5}\)

g) \(\dfrac{5}{3}:\dfrac{5}{-3}=\dfrac{5}{3}:\dfrac{-5}{3}=\dfrac{5}{3}\cdot\dfrac{-3}{5}=-1\)

h) \(\dfrac{5}{4}:\left(-9\right)=\dfrac{5}{4}:\dfrac{-9}{1}=\dfrac{5}{4}\cdot\dfrac{-1}{9}=-\dfrac{5}{36}\)