\(\dfrac{5}{6}=\dfrac{10}{12}=\dfrac{30}{36}\)

\(\dfrac{9}{12}=\dfrac{6}{8}=\dfrac{15}{20}\)

a)5/6=10/12=30/36

lm đc mỗi phần a,ko bt giải thik

a.\(\dfrac{5}{6}=\dfrac{10}{12}=\dfrac{30}{36}\)

b.\(\dfrac{9}{12}=\dfrac{6}{8}=\dfrac{15}{20}\)

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

Áp dụng công thức: \(1+2+3+...+n=\dfrac{n+\left(n+1\right)}{2}\) ta có:

\(A=\dfrac{2}{2.3}+\dfrac{2}{4.5}+...+\dfrac{2}{98.99}=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=\dfrac{64}{99}< \dfrac{66}{99}=\dfrac{2}{3}\)

Ta có: \(\dfrac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

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

\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{10}+\sqrt{2}}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{10}+\sqrt{2}}\)

\(=\dfrac{6+2\sqrt{5}}{3\sqrt{2}+\sqrt{10}}-\dfrac{6-2\sqrt{5}}{3\sqrt{2}-\sqrt{10}}\)

\(=\dfrac{\left(6+2\sqrt{5}\right)\left(3\sqrt{2}-\sqrt{10}\right)-\left(6-2\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)}{8}\)

\(=\dfrac{18\sqrt{2}-6\sqrt{10}+6\sqrt{10}-10\sqrt{2}-18\sqrt{2}-6\sqrt{10}+6\sqrt{10}+10\sqrt{2}}{8}\)

\(=0\)

a) \(\dfrac{2}{5}\times?=\dfrac{3}{10}\)

\(?=\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{4}\)

b) \(\dfrac{1}{8}:?=\dfrac{1}{5}\)

\(?=\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)

a: Phân số cần tìm là: \(\dfrac{3}{10}:\dfrac{2}{5}=\dfrac{3}{10}\cdot\dfrac{5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\)

b: Phân số cần tìm là \(\dfrac{1}{8}:\dfrac{1}{5}=\dfrac{5}{8}\)

\(\dfrac{1}{5} = \dfrac{{1.3}}{{5.3}} = \dfrac{3}{{15}}\); 

\(\dfrac{{ - 10}}{{55}} = \dfrac{{ - 10:5}}{{55:5}} = \dfrac{{ - 2}}{{11}}\)

Vậy các cặp phân số bằng nhau là: \(\dfrac{1}{5} = \dfrac{3}{{15}}; \dfrac{{ - 10}}{{55}} = \dfrac{{ - 2}}{{11}}\)

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

Lời giải:

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

cách 1:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{6}{1}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{5}{1}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{3}{1}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{18}{3}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{15}{3}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{9}{3}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{16}{3}+\dfrac{1}{2}\right)-\left(\dfrac{20}{3}-\dfrac{3}{2}\right)-\left(\dfrac{2}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{32}{6}+\dfrac{3}{6}\right)-\left(\dfrac{40}{6}-\dfrac{9}{6}\right)-\left(\dfrac{4}{6}+\dfrac{15}{6}\right)\)

= \(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

= \(-\dfrac{15}{6}=-\dfrac{5}{2}\)

cách 2:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

= \(\left(-2\right)-0+\dfrac{1}{2}\)

= \(-\dfrac{5}{2}\)

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

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

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

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

\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

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

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

=-2

4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)

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

=4