\(\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
Giúp mình tính bài này nha!!! Mình đang cần gấp ạ
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
Giúp mình bài này với mình đang cần gấp
Rút gọn các biểu thức sau:
C=\(\frac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)\(-\frac{\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
D=\(\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
\(\sqrt{5-\sqrt{6}+\sqrt{10}-\sqrt{15}}-\frac{\sqrt{16-4\sqrt{15}}}{2}\)
\(\frac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}}-4}\)
\(\sqrt{3+\sqrt{2}+\sqrt{3}+\sqrt{6}}-\sqrt{2+\sqrt{3}}\)
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Giải dùm mình nhé!!!!!!! Không cần giải hết đâu!!!!! Mọi người thấy bài nào dạng giống nhau thì giải 1 lần rồi nói mình đó là bài nào lun nha!!!!!!!!!! Mình cần rất gấp ạ!!!!!! Mong các bạn giúp mình
có thể giúp mình giải bài này với đc k ạ mình đang cần gấp (xin cảm ơn)
Bài 1:
a,\(3x-7\sqrt{x}+4=0\)
b, \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}-4}=\dfrac{6-\sqrt{x}}{7-\sqrt{x}}\)
d, \(\sqrt{x-3}-\dfrac{5}{3}\sqrt{9x-27}+\dfrac{3}{2}\sqrt{4x-12}=-1\)
Bài 2:
a, \(\sqrt{x^2+6x+9}=3x-6\)
b, \(\sqrt{3x^2}=x+2\)
c, \(\sqrt{x^2-4x+4}-2x+5=0\)
d, \(x^2-2\sqrt{7x}+7=0\)
Bài 3:
a, \(\sqrt{3+x}+\sqrt{6-x}=3\)
b, \(\sqrt{3+x}-\sqrt{2-x}=1\)
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
Bài 1
a, `3x-7\sqrt{x}+4=0` ĐKXĐ : `x>=0`
`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`
`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`
`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`
TH1 :
`3\sqrt{x}-4=0`
`<=>\sqrt{x}=4/3`
`<=>x=16/9` ( tm )
TH2
`\sqrt{x}-1=0`
`<=>\sqrt{x}=1` (tm)
Vậy `S={16/9;1}`
b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17` ĐKXĐ : `x>=1`
`<=>(1/2-9/2+3)\sqrt{x-1}=-17`
`<=>-\sqrt{x-1}=-17`
`<=>\sqrt{x-1}=17`
`<=>x-1=289`
`<=>x=290` ( tm )
Vậy `S={290}`
Bài 1:
a) Ta có: \(3x-7\sqrt{x}+4=0\)
\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Mọi người ơi giúp mình với. Mình đang cần gấp
\(\sqrt{5}+3\sqrt{3}+\sqrt{6-3\sqrt{3}}\) =?
A.\(4\sqrt{2}\) B.\(\sqrt{2}\) C.3\(\sqrt{2}\) D.2\(\sqrt{2}\)
\(\text{Theo đề bài: }=\dfrac{3\sqrt{2}+6\sqrt{3}+2\sqrt{5}-\sqrt{6}}{2}\)
Giúp mình tính bài này với ạ =)) cần gấp.
a) \(\left(2+\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
b) \(\sqrt{6-2\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}\)
c) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
d) \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}-3\sqrt{2}\)
c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)
\(=5+3\sqrt{2}\)
d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\left(\sqrt{10}-\sqrt{2}\right)\)
mình đang cần gấp giải giúp mình nha
a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}+\sqrt{2}\right)\)
\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)
\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)
b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)
\(=2.\left(5-1\right)=2.4=8\)
1) Tính:
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5-1}}\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\)
Giúp mình với, mình cần gấp
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
Câu b á bạn, chỗ \(\dfrac{4}{\sqrt{5-1}}\) là đề như vậy hay là \(\dfrac{4}{\sqrt{5}-1}\) vậy?
Tính:
\(2\sqrt{4+\sqrt{6-2\sqrt{5.\left(\sqrt{10}-\sqrt{2}\right)}}}\)
Mình cần gấp ạ