help

11/18

THAM KHẢO:

a) 9
b)90
c)8
d)7/10
e)2/5

a=9

b=90

c=2

d=0,7

e=0,2

`@` `\text {Ans}`

`\downarrow`

\(S=\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)

`S=0,7 + 1/3 - 5/2`

`S=31/30 - 5/2 = -22/15`

Bài 1: 

a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)

b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)

Bài 2: 

a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)

b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)

c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)

\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)

\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)

\(=\dfrac{6}{35}\)

a: \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)

b: \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)

a) \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)

b) \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)

\(A=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)

=10-1

=9

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

b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)

\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)

\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)

a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)

\(=-\dfrac{4}{3}\cdot0.4\)

\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)

b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)

\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)

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

\(=\dfrac{6}{7}\)

1.

Đặt biểu thức là $A$

Ta thấy:

$\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$

Tương tự với các phân số còn lại và công theo vế thì:

$A=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+...+(\sqrt{2019}-\sqrt{2018})$

$=\sqrt{2019}-1$

2.

$\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{5.3}+3}+\sqrt{3-2\sqrt{3.1}+1}$

$=\sqrt{(\sqrt{5}-\sqrt{3})^2}+\sqrt{(\sqrt{3}-1)^2}$

$=|\sqrt{5}-\sqrt{3}|+|\sqrt{3}-1|$

$=\sqrt{5}-\sqrt{3}+\sqrt{3}-1=\sqrt{5}-1$

a: \(\sqrt{4\cdot36}=\sqrt{144}=12\)

b: \(\left(\sqrt{8}-3\sqrt{2}\right)\cdot\sqrt{2}\)

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

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

c: \(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}=\dfrac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=-\sqrt{7}\)

d: \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{\sqrt{5}-2}\)

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

\(=2\sqrt{5}-4+2\sqrt{5}+4=4\sqrt{5}\)