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8 tháng 9 2021
\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\\ =\sqrt{15-6\sqrt{6}}+\sqrt{\left(2\sqrt{6}-3\right)^2}\\ =\sqrt{\left(3-\sqrt{6}\right)^2}+2\sqrt{6}-3\\ =3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
\(c,\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\\ =\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}}\\ =\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =3-\sqrt{3}+\sqrt{3}-1=2\)
8 tháng 9 2021
c: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1
=2
B
1
12 tháng 9 2021
Ta có: \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\sqrt{\left(2-\sqrt{5}\right)^2}-2\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}-2-2\sqrt{5}\)
=-2
0H
1
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2022
ĐKXĐ: \(x>2\)
\(A=\dfrac{\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2+4\sqrt{x-2}+4}}{\sqrt{\left(\dfrac{2}{x}-1\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}}{\left|\dfrac{2}{x}-1\right|}=\dfrac{\left|\sqrt{x-2}-2\right|+\left|\sqrt{x+2}+2\right|}{1-\dfrac{2}{x}}\)
- Với \(x\ge6\Rightarrow A=\dfrac{\sqrt{x-2}-2+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{2x\sqrt{x-2}}{x-2}=\dfrac{2x}{\sqrt{x-2}}\)
- Với \(2< x< 6\Rightarrow A=\dfrac{2-\sqrt{x-2}+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{4x}{x-2}\)
9 tháng 11 2021
\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
30 tháng 7 2018
\(C=\sqrt{\frac{x-2\sqrt{xy}+y}{x+6\sqrt{xy}+y}}\)
\(C=\sqrt{\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.\sqrt{y}+\left(\sqrt{y}\right)^2}{\left(\sqrt{x}\right)^2+2\sqrt{x}\sqrt{y}+\left(\sqrt{y}\right)^2+4\sqrt{xy}}}\)
\(C=\sqrt{\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2+4xy}}\)
19 tháng 7 2021
a) Ta có: \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b) Để B=16 thì \(4\sqrt{x+1}=16\)
\(\Leftrightarrow x+1=16\)
hay x=15
9 tháng 9 2021
\(C=\left(\dfrac{3}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
HA
7 tháng 9 2015
\(\frac{a\left(a-1\right)}{a-1}-\frac{a\left(-6a+5\right)}{a-1}=\frac{a^2-a+6a^2-5a}{a-1}\)
=\(\frac{7a^2-6a}{a-1}\)
23 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{5x+10\sqrt{x}-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5x+10\sqrt{x}-5\sqrt{x}+6+x-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
23 tháng 9 2021
\(P=\dfrac{5\sqrt{x}}{\sqrt{x}-2}-\dfrac{3-\sqrt{x}}{\sqrt{x}+2}+\dfrac{6x}{4-x}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+2\right)-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{5x+10\sqrt{x}+x-5\sqrt{x}+6-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{5\sqrt{x}+6}{x-4}\)
\(e,\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\\ =\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\\ =\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(f,\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\\ =36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}=36-36\sqrt{2}+27\sqrt{3}\)
\(g,\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{2}\right)\sqrt{3-2\sqrt{2}}\\ =\left[\left(2-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\right]\left(3+\sqrt{2}\right)\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left(3-4\sqrt{2}\right)\left(3+\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =\left(1-9\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =10\sqrt{2}-37\)
\(h,A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\\ A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)\\ A^2=6+2\sqrt{5}\\ A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)