a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

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

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

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

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

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

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

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

\(P=2\)

b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)

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

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

\(N=-4\)

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

\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)

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

\(Q=\left(\sqrt{5}\right)^2-2^2\)

\(Q=1\)

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

Lời giải chi tiết: 

5-1=4     4-1=3    3-1=2   2+3=5     

5-2=3     4-2=2    3-2=1   3+2=5

5-3=2     4-3=1    2-1=1   5-2=3

5-4=1                               5-3=2

Tính:

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

\(7832\)

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

Lời giải chi tiết:

1)

\(\dfrac{5}{\sqrt{5}}=\dfrac{5\sqrt{5}}{5}\sqrt{5}\)

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

\(\dfrac{5}{\sqrt{7}}=\dfrac{5\sqrt{7}}{\sqrt{49}}=\left(\dfrac{5}{7}\right)\sqrt{7}\)

1:

\(\dfrac{2\sqrt{3}}{5\sqrt{7}}=\dfrac{2\sqrt{21}}{35}\)

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

2: \(\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

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

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

\(\dfrac{12}{\sqrt{5}-\sqrt{3}}=6\left(\sqrt{5}+\sqrt{3}\right)=6\sqrt{5}+6\sqrt{3}\)

Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)

Bài 3:

a, 800 kg 5g > 8,005 kg

b, 9 ha 4 dam² < 9\(\dfrac{5}{100}\) ha

c, 5m 5mm > 5,0005 m

d, 250 kg > \(\dfrac{1}{5}\) tấn

5 - 1 = 4     1 + 4 = 5     2 + 3 = 5

5 - 2 = 3     4 + 1 = 5     3 + 2 = 5

5 - 3 = 2     5 - 1 = 4     5 - 2 = 3

5 - 4 = 1     5 - 4 = 1     5 - 3 = 2

5-1=4     1+4=5     2+3=5

5-2=3     4+1=4     3+2=5

5-3=2     5-1=4      5-2=3

5-4=1     5-4=1      5-3=2

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)