Rút gọn: a, \(A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
b, \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
bài 1 rút gọn
a \(A=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
b\(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
c\(C=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) d\(D=\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`
`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`
`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`
`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`
`=sqrt2(sqrt5+1+sqrt5-1)`
`=sqrt{2}.2sqrt5`
`=2sqrt{10}`
`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`
`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`
`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`
`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`
`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`
`=2(7-3)`
`=8`
`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`
`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`
`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`
`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`
`=2/sqrt2=sqrt2`
`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`
`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`
`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`
`=>sqrt2D=8`
`=>D=4sqrt2`
RÚT GỌN BIỂU THỨC
A= \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\)\(\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
B= \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\)\(\left(\sqrt{6}+11\right)\)
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Câu 3: Rút gọn biểu thức sau:
a. \(\dfrac{1}{\sqrt{5}-1}+\dfrac{1}{1+\sqrt{5}}\)
b. \(\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
c. \(\dfrac{2}{\sqrt{5}+\sqrt{3}}-\dfrac{3-\sqrt{15}}{\sqrt{5}-\sqrt{3}}\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
Dạng 3.Chứng minh đẳng thức
Bài 1: CM
a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
b)\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Bài 2 :CM
\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{2}}=\sqrt{\sqrt{5}+1}\)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
* Rút gọn biểu thức
a. \(\left(2\sqrt{125}-3\sqrt{5}-\sqrt{180}\right):\left(-\sqrt{5}\right)+\sqrt{8}\)
b. \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
c. \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
d.\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}\right)\)
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
rút gọn
g, \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right).\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\) h,\(\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\dfrac{1}{3}}\right).\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\dfrac{1}{5}}\right)\)
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
Thực hiện phép tính và thu gọn biểu thức:
B= \(\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
Thực hiện phép tính:
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
1) Rút gọn:
a) A = \(\sqrt{5-2\sqrt{3-\sqrt{3}}}-\sqrt{3+\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) B = \(\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}+\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}\)
c) C = \(\dfrac{\sqrt{21+3\sqrt{5}}+\sqrt{21-3\sqrt{5}}}{\sqrt{21}+6\sqrt{11}}+\sqrt{11-6\sqrt{2}}\)
d) D = \(\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right).\sqrt{\dfrac{2+2\sqrt{5}}{2+\sqrt{5}}}\)
e) E = \(\dfrac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)