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2 tháng 2 2021

\(E=(\sqrt{18}-3\sqrt{6}+\sqrt{2}).\sqrt{2}+6\sqrt{3} \\ = (3\sqrt{2}-3\sqrt{6}+\sqrt{2}).\sqrt{2} + 6\sqrt{3} \\ = 6 - 6\sqrt{3}+2 + 6\sqrt{3} \\ = 8\)

 

2 tháng 2 2021

\(G=(2\sqrt2-\sqrt5+\sqrt{18}).(\sqrt{50}+\sqrt5) \\ =(2\sqrt2-\sqrt5+3\sqrt2).\sqrt5(\sqrt{10}+1) \\ = (5\sqrt2-\sqrt5). \sqrt5 (\sqrt{10}+1) \\ = (5\sqrt{10}-5)(\sqrt{10}+1) \\ = 5(\sqrt{10}-1)(\sqrt{10}+1)=5.9=45\)

31 tháng 12 2023

Bài 1:

ĐKXĐ: \(\dfrac{5}{x^2+6}>=0\)

=>\(x^2+6>0\)

mà \(x^2+6>=6>0\forall x\)

nên \(x\in R\)

Bài 2:

a: Sửa đề: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\cdot\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)

b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{6}\)

\(=\dfrac{\sqrt{12}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\)

e: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=\sqrt[3]{-27}=-3\)

f: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)

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

30 tháng 6 2021

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

\(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

\(-33\sqrt{2}\)

30 tháng 6 2021

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

\(10-2\sqrt{21}+14\sqrt{21}\)

\(10+12\sqrt{21}\)

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

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

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

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

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

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

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

bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17

6 tháng 8 2021

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

25 tháng 9 2021

a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)

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

d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)