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\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)
\(B=\sqrt{5}+2+\sqrt{5}-2\)
\(B=2\sqrt{5}\)
\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)
\(A=-1\)
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) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{12-2\cdot2\sqrt{3}+1}-\sqrt{18+4\cdot3\sqrt{2}+4}\)
\(=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)
\(=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2\)
\(=-3\)
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
2:
a: =căn 17-4-căn 17=-4
b: =5-2căn 3-2căn 3=5-4căn 3
1:
a: =>|x+1|=-x
=>x<=0 và (x+1)^2=x^2
=>x<=0 và (x+1+x)(x+1-x)=0
=>x=-1/2
a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)
\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)
\(=36\sqrt{1-a^2}\)
c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)
\(=15a-3a=12a\)
b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)
\(=a^2\)
d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
\(=a^2-6a+9-\sqrt{36a^2}\)
\(=a^2-6a+9-\left|6a\right|\)
\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)
`A=\sqrt{6-2\sqrt{5}}`
`A=\sqrt{(\sqrt{5}-1)^2}`
`A=\sqrt{5}-1`
_________
`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`
`B=\sqrt{(\sqrt{3}-1)^2}`
`B=\sqrt{3}-1`
_________
`C=\sqrt{19-8\sqrt{3}}`
`C=\sqrt{(4-\sqrt{3})^2}`
`C=4-\sqrt{3}`
_________
`D=\sqrt{5-2\sqrt{6}}`
`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`
`D=\sqrt{3}-\sqrt{2}`
\(A=\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}=\sqrt{ \left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)
\(D=\sqrt{5-2\sqrt{6}}=\sqrt{5-2.\sqrt{2}.\sqrt{3}}=\sqrt{\sqrt{3}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{2^2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
\(A=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
\(=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+\sqrt{1^2}}-\sqrt{\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}+2^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\)
\(=-2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-1-3\sqrt{2}-2\)
\(=-3\)
\(A=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2=-3\)