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a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}=\sqrt{5}.\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\sqrt{5}.\left(1-\sqrt{2}\right)=\sqrt{5}-\sqrt{5}.\sqrt{2}=\sqrt{5}-\sqrt{10}\)
b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}=\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\sqrt{27}.\left(2-\sqrt{5}\right)=2\sqrt{27}-\sqrt{135}\)
c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}=\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)
\(=\dfrac{\sqrt{2}}{3-\sqrt{10}}\)
d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}=\dfrac{\sqrt{5\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)
\(=\dfrac{\sqrt{5}.\left(1-\sqrt{3}\right)}{2}=\dfrac{\sqrt{5}-\sqrt{15}}{2}\)
Chúc bạn học tốt!!!
a) \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)
= \(\sqrt{5}.\sqrt{\left(1-\sqrt{2}\right)^2}\)
= \(\sqrt{5}.\left(\sqrt{2}-1\right)\)
= \(\sqrt{10}-\sqrt{5}\)
b) \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)
= \(\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)
= \(\sqrt{27}.\left(\sqrt{5}-2\right)\)
= \(\sqrt{135}-2\sqrt{27}\)
c) \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)
= \(\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)
= \(\dfrac{\sqrt{2}}{\sqrt{10}-3}\)
d) \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)
= \(\dfrac{\sqrt{5}.\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)
= \(\dfrac{\sqrt{5}.\left(\sqrt{3}-1\right)}{2}\)
= \(\dfrac{\sqrt{15}-\sqrt{5}}{2}\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
Câu 1 :
a ) \(\sqrt{0,36.100}=\sqrt{36}=6\)
b ) \(\sqrt[3]{-0,008}=\sqrt[3]{\left(-0,2\right)^3}=-0,2\)
c ) \(\sqrt{12}+6\sqrt{3}+\sqrt{27}=2\sqrt{3}+6\sqrt{3}+3\sqrt{3}=11\sqrt{3}\)
Câu 2 :
a ) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a-\sqrt{ab}+b\)
\(\sqrt{96}.\sqrt{125}\)
\(\sqrt{16.6}\sqrt{25.5}\)
\(4.5\sqrt{6.5}\)
\(20\sqrt{30}\)
\(b,\sqrt{a^4b^5}\)
\(a^2b^2\sqrt{b}\)
\(c,\sqrt{a^6b^{11}}\)
\(a^3b^5\sqrt{b}\)
\(d,\sqrt{a^3\left(1-a\right)^4}\)
\(a\left(1-a\right)^2\sqrt{a}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
a: \(=\dfrac{10}{9}\left(\dfrac{2}{5}\sqrt{5}+\dfrac{1}{2}\sqrt{5}\right)=\dfrac{10}{9}\cdot\dfrac{9}{10}\sqrt{5}=\sqrt{5}\)
b: \(=\dfrac{4}{3}\sqrt{2}+\sqrt{2}+\dfrac{1}{6}\sqrt{2}=\dfrac{5}{2}\sqrt{2}\)
c: \(=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
d: \(=6\sqrt{a}+\dfrac{2}{3}\cdot\dfrac{1}{2}\sqrt{a}-3\sqrt{a}+7=\dfrac{10}{3}\sqrt{a}+7\)
Bài 4:
a: ĐKXĐ: x>=0; x<>1
b: \(P=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{2a^2+4}{-\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{-\sqrt{a}+1+\sqrt{a}+1}{a-1}\)
\(=\dfrac{-2a^2-4}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{2}{a-1}\)
\(=\dfrac{-2a^2-4+2a^2+2a+2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{2a+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
a) \(\sqrt{27\left(9-4\sqrt{5}\right)}=3\sqrt{3\left(\sqrt{5}-2\right)^2}=3\sqrt{3}\left(\sqrt{5}-2\right)=3\sqrt{15}-6\sqrt{3}\)
b) \(\sqrt{a^4b^5}=a^2b^2\sqrt{b}\)
c) \(\sqrt{a^3\left(1-a\right)^4}=a\left(1-a\right)^2\sqrt{a}\)
d) không biết