a, =2.0,1-0,5                                  b,=\(\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

    =0,2-0,5                                        =  -10:(\(\dfrac{-5}{7}\))

    =-0,3                                             = 14

           c, \(=\dfrac{5^4.\left(4.5\right)^4}{\left(5^2\right)^4.4^5}=\dfrac{5^4.4^4.5^4}{5^8.4^5}=\dfrac{1}{4^{ }}\)

1: \(=\left(\dfrac{1}{3}\right)^{25}\cdot90^{25}\cdot\dfrac{1}{3^{25}}-\dfrac{2}{12}\)

\(=\dfrac{30^{25}}{3^{25}}-\dfrac{1}{6}=10^{25}-\dfrac{1}{6}\)

2: \(=10\cdot1\cdot\dfrac{4}{3}+7-\dfrac{1}{6}\cdot2=20\)

1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)

2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)

3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)

4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)

5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)

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

\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)

\(=0\cdot\sqrt{6}=0\)

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)

\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)

=1-1/5=4/5

Với `n` làm cho biểu thức dưới đây có nghĩa, ta có:

`1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn sqrt(n+1)(sqrt(n+1)+sqrt(n)))=(sqrt(n+1)-sqrt(n))/(sqrtn sqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)`

Khi đó:
`M=\sum_{n=1}^(24)=1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)=1/(sqrt1)-1/(sqrt25)=1-1/5=4/5`

1.

0,2 . \(\sqrt{100}\) - \(\sqrt{\dfrac{16}{25}}\)

= 0,2 . 10 - \(\dfrac{4}{5}\)

= 2 - \(\dfrac{4}{5}\)

= \(\dfrac{6}{5}\)

1/ \(0,2.\sqrt{100}-\sqrt{\dfrac{16}{25}}\)

\(=0,2.10-0,8\)

\(=2-0,8=1,2\)

2/ \(\dfrac{2^7.9^3}{6^5.8^2}\)

\(=\dfrac{93312}{497664}=\dfrac{3}{16}=0,1875\)

3/ \(\sqrt{0,01}-\sqrt{0,25}\)

\(=0,1-0,5\)

\(=-0,4\)

4/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)

\(=0,5.10-0,5\)

\(=5-0,5=4,5\)

5/ \(7.\sqrt{0,01}+2.\sqrt{0,25}\)

\(=7.0,1+2.0,5\)

\(=0,7+1=1,7\)

6/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)

\(=0,5.10-0,2\)

\(=5-0,2=4,8\)

Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

Do đó :

S\(< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)\)\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)(dpcm)

`#3107.101107`

a)

`2/5 \sqrt{25} - 1/2 \sqrt{4}`

`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`

`= 2/5*5 - 1/2*2`

`= 2 - 1`

`= 1`

b)

`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`

`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`

`= 0,5*0,3 + 5*0,9`

`= 0,15 + 4,5`

`= 4,65`

c)

`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`

`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`

`= 2/5*5/6 - 5/2*2/5`

`= 1/3 - 1`

`= -2/3`

d)

`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`

`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`

`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`

`= -2*6/4 + 5*9/5`

`= -3 + 9`

`= 6`

a: \(\dfrac{1}{\sqrt{5}-\sqrt{3}+2}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{\left(\sqrt{5}-\sqrt{3}\right)^2-4}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{8-2\sqrt{15}-4}=\dfrac{\sqrt{5}-\sqrt{3}-2}{4-2\sqrt{15}}\)

\(=\dfrac{\left(\sqrt{5}-\sqrt{3}-2\right)\left(4+2\sqrt{15}\right)}{16-60}\)

\(=\dfrac{4\sqrt{5}+2\cdot\sqrt{75}-4\sqrt{3}-2\sqrt{45}-8-4\sqrt{15}}{-44}\)

\(=\dfrac{-2\sqrt{5}+6\sqrt{3}-8-4\sqrt{15}}{-44}\)

\(=\dfrac{\sqrt{5}-3\sqrt{3}+4+2\sqrt{15}}{22}\)

b: Sửa đề: \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)

\(=\dfrac{-1+\sqrt{2}}{2-1}+\dfrac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\dfrac{-\sqrt{24}+\sqrt{25}}{25-24}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}+...+\left(-\sqrt{24}\right)+\sqrt{25}\)

=5-1

=4