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11 tháng 7 2017

\(=\sqrt{\dfrac{1.32.225}{125.35.56}}\\ =\sqrt{\dfrac{2^5.25.9}{25.5.35.8.7}}\\ =\sqrt{\dfrac{2^2.9}{5.35.7}}\\ =\sqrt{\dfrac{36}{1225}}\\ =\dfrac{6}{35}\)

11 tháng 7 2017

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

\(=\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}.\sqrt{\dfrac{225}{56}}\)

\(=\sqrt{\dfrac{1.32.225}{125.35.56}}=\sqrt{\dfrac{2^5.5^2.3^2}{5^3.5.7.7.2^3}}\)

\(=\sqrt{\dfrac{2^2.3^2}{5^2.7^2}}=\dfrac{6}{35}\)

Chúc bạn học tốt!!!

25 tháng 8 2017

\(\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}=\sqrt{\dfrac{1}{125}.\dfrac{32}{35}:\dfrac{56}{225}}=\sqrt{\dfrac{36}{1225}}=\dfrac{6}{35}\)

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

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

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

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

13 tháng 7 2021

em cảm ơn ạ yeu

24 tháng 8 2023

\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\) 

\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)

\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)

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

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

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

\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)

\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)

\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)

17 tháng 1 2022

a) 2√6>3√2>√13>2√326

b)1/3√39>1/4√32>1/5√35>1/2√51339

@@@

17 tháng 1 2022

Bạn Tạ Bảo Trân làm sai

27 tháng 10 2021

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

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

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

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

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

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

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

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

15 tháng 8 2021

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

15 tháng 8 2021

mik cảm ơn ạ