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\(\sqrt{8,1}.\sqrt{250}\)

\(=\sqrt{81}.\sqrt{25}\)

\(=9.5\)

\(=45\)

\(\sqrt{2,5}.\sqrt{360}\)

\(=\sqrt{25}.\sqrt{36}\)

\(=5.6\)

\(=30\)

\(\sqrt{\frac{-49}{-121}}=\sqrt{\frac{49}{121}}\)

\(=\frac{\sqrt{49}}{\sqrt{121}}\)

\(=\frac{7}{11}\)

\(\sqrt{\frac{-36}{-169}}=\sqrt{\frac{36}{169}}\)

\(=\frac{\sqrt{36}}{\sqrt{169}}=\frac{6}{13}\)

26 tháng 7 2018

\(\sqrt{2,5}.\sqrt{360}\)

\(=\sqrt{25}.\sqrt{36}\)

\(=5.6\)

\(=30\)

\(\sqrt{\frac{-49}{-121}}\)

\(=\sqrt{\frac{49}{121}}\)

\(=\frac{\sqrt{49}}{\sqrt{121}}=\frac{7}{11}\)

18 tháng 10 2020

a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)

\(=6.11+\left(-4\right)-5=66-9=57\)

b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)

\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)

\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)

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

c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)

\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)

\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)

\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)

a) Ta có: \(\sqrt{45}:\sqrt{80}\)

\(=\sqrt{\frac{45}{80}}=\sqrt{\frac{9}{20}}\)

\(=\frac{3}{2\sqrt{5}}\)

b) Ta có: \(\sqrt{\frac{3}{15}}:\sqrt{\frac{36}{45}}\)

\(=\sqrt{\frac{1}{5}:\frac{4}{5}}\)

\(=\sqrt{\frac{1}{5}\cdot\frac{5}{4}}\)

\(=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

c) Ta có: \(\sqrt{\frac{72}{9}}:\sqrt{8}\)

\(=\frac{\sqrt{8}}{\sqrt{8}}=1\)

d) Ta có: \(\sqrt{\frac{288}{169}}:\sqrt{\frac{8}{225}}\)

\(=\sqrt{\frac{288}{169}:\frac{8}{225}}\)

\(=\sqrt{\frac{288}{169}\cdot\frac{225}{8}}\)

\(=\sqrt{\frac{8100}{169}}=\frac{90}{13}\)

25 tháng 8 2017

tu lam di cau nao kho thi hoi hoi vay ko ai tra loi cho dau

cau e)

\(A=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)(suy ra A>=0)

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

\(A^2=1\)

A=1

(bai toan co nhieu cach)

cau m)

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

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

\(=1\)

cau G)

\(=\frac{5\sqrt{7}}{\sqrt{35}}-\frac{7\sqrt{5}}{\sqrt{35}}+\frac{2\sqrt{70}}{\sqrt{35}}\)

\(=\frac{5}{\sqrt{5}}-\frac{7}{\sqrt{7}}+2\sqrt{2}\)

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

25 tháng 6 2019

1) \(\sqrt{\frac{24}{3}}\cdot\sqrt{\frac{3a}{8}}=\sqrt{\frac{72a}{24}}=\sqrt{3a}\)

2) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{676}=26\)

3) \(\sqrt{5a}\cdot\sqrt{45a}-3a=\sqrt{225a^2}-3a=15a-3a=12a\)

4) \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}=a^2-6a+9-\sqrt{36a^2}=a^2-6a+9-6a=a^2-12a+9\)

26 tháng 11 2016

 căn bậc 3 x căn bận 12=6

cân bậc 12 chia căn bâc 3 =2

20 tháng 6 2019

ĐKXĐ: \(x\ge0\)

Từ biểu thức đầu suy ra: \(7\left(5\sqrt{x}-2\right)=2\left(8\sqrt{x}+2,5\right)\)

\(35\sqrt{x}-14=16\sqrt{x}+5\)

\(19\sqrt{x}=19\Rightarrow x=1\)(Thỏa mãn ĐKXĐ)