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\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)
\(=\sqrt{8-2\cdot\sqrt{8}\cdot2+4}+2\sqrt{2}+1\)
=2căn 2-2+2căn 2+1
=4căn 2-1
\(\dfrac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}=\dfrac{\sqrt{\left(\sqrt{5}-2\right)^2}}{2-\sqrt{5}}=\dfrac{\sqrt{5}-2}{2-\sqrt{5}}=\dfrac{-\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=-1\)
\(\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2}-1}=\sqrt{9}=3\)
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)=\(\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)
=\(\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)=\(\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\sqrt{13+30\sqrt{2}+30}\)
=\(\sqrt{43+30\sqrt{2}}\)=\(\sqrt{\left(5+3\sqrt{2}\right)^2}\)=\(5+3\sqrt{2}\)
1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}+2-\sqrt{5}\)
\(=4\)
2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=3-\sqrt{3}+3+\sqrt{3}\)
\(=6\)
`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`
Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`
`b)` Với `x >= 0,x ne 4` có:
`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[\sqrt{x}-2]/[\sqrt{x}-3]`
`c)` Với `x >= 0,x ne 4` có:
`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`
Có: `C >= 1`
`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`
`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`
`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`
Vì `x >= 0=>\sqrt{x} >= 0`
`=>\sqrt{x}-3 > 0`
`<=>x > 9` (t/m đk)
a: ĐKXĐ: a>=0; a<>4
b: \(M=\dfrac{a\sqrt{a}-a\sqrt{a}+2a-a-2\sqrt{a}}{a-4}=\dfrac{a-2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}}{\sqrt{a}+2}\)
c: Khi a=9 thì \(M=\dfrac{3}{3+2}=\dfrac{3}{5}\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}.\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{13+30\sqrt{2}+30}=\sqrt{43+30\sqrt{2}}\)
\(\sqrt{\dfrac{9}{4}}-\sqrt{2}+\sqrt{2}\\ =\dfrac{3}{2}-\left(\sqrt{2}-\sqrt{2}\right)\\ =\dfrac{3}{2}-0\\ =\dfrac{3}{2}\)