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`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{2}}{2}\)
___________
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
__________
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)
1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
2)
a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:
\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)
b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
=1
Thay x=1 vào P, ta được:
\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)
a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{2}=\dfrac{4}{2}=2\)
b: \(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{15}}=\sqrt{\dfrac{6}{15}}=\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}\)
c: \(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\)
\(=3\sqrt{a}+9\sqrt{a}+3\cdot5\sqrt{a}-16\cdot7\sqrt{a}\)
\(=27\sqrt{a}-112\sqrt{a}=-85\sqrt{a}\)
d: \(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\sqrt{ab}-\sqrt{bc}}\)
\(=\sqrt{ab}+\sqrt{bc}\)
e: \(a\left(\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\cdot\sqrt{\dfrac{a}{b}}}\right)\cdot\sqrt{ab}\)
\(=a\cdot\sqrt{\dfrac{a}{b}\cdot ab+2\sqrt{ab}\cdot ab+b\cdot\sqrt{\dfrac{a}{b}}\cdot ab}\)
\(=a\cdot\sqrt{a^2+2\cdot ab\cdot\sqrt{ab}+a\sqrt{a}\cdot b\sqrt{b}}\)
\(=a\cdot\sqrt{a^2+3\cdot a\cdot\sqrt{a}\cdot b\cdot\sqrt{b}}\)
e: ĐKXĐ: a>=0 và a<>1
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\left(a-\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)\)
a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=-4\sqrt{5}+15\sqrt{2}\)
b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)
\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)
\(=8\sqrt{3}+2\sqrt{2}-4\)
c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3
=6
d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
Ta có: \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
Ta có: \(a=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
=2
Thay a=2 và \(b=\dfrac{2}{3}\) vào M, ta được:
\(M=\dfrac{1+2\cdot\dfrac{2}{3}}{2+\dfrac{2}{3}}-\dfrac{1-2\cdot\dfrac{2}{3}}{2-\dfrac{2}{3}}\)
\(=\dfrac{7}{8}+\dfrac{1}{4}\)
\(=\dfrac{7}{8}+\dfrac{2}{8}=\dfrac{9}{8}\)