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a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-3}{2}\)

17 tháng 12 2023

\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)

\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)

\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}=-2\)

\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)

17 tháng 5 2021
) V T = ( 2 √ 3 − √ 6 √ 8 − 2 − √ 216 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 2 ⋅ √ 3 − √ 6 √ 2 2 ⋅ 2 − 2 − √ 6 2 .6 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 6 − √ 6 2 √ 2 − 2 − 6 . √ 6 3 ) ⋅ 1 √ 6 = [ √ 6 ( √ 2 − 1 ) 2 ( √ 2 − 1 ) − 6 √ 6 3 ] ⋅ 1 √ 6 = ( √ 6 2 − 2 √ 6 ) ⋅ 1 √ 6 = ( √ 6 2 − 4 √ 6 2 ) ⋅ 1 √ 6 = ( − 3 2 √ 6 ) ⋅ 1 √ 6 = − 3 2 = − 1 , 5 = V P . b) V T = ( √ 14 − √ 7 1 − √ 2 + √ 15 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = ( √ 7 ⋅ √ 2 − √ 7 1 − √ 2 + √ 5 ⋅ √ 3 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = [ √ 7 ( √ 2 − 1 ) 1 − √ 2 + √ 5 ( √ 3 − 1 ) 1 − √ 3 ] : 1 √ 7 − √ 5 = ( − √ 7 − √ 5 ) ( √ 7 − √ 5 ) = − ( √ 7 + √ 5 ) ( √ 7 − √ 5 ) = − ( 7 − 5 ) = − 2 = V P . c) V T = a √ b + b √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a ⋅ √ b + √ b ⋅ √ b ⋅ √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a b + √ b ⋅ √ a b √ a b : 1 √ a − √ b = √ a b ( √ a + √ b ) √ a b ⋅ ( √ a − √ b ) = ( √ a + √ b ) ⋅ ( √ a − √ b ) = a − b = V P . d) V T = ( 1 + a + √ a √ a + 1 ) ( 1 − a − √ a √ a − 1 ) = ( 1 + √ a ⋅ √ a + √ a √ a + 1 ) ( 1 − √ a ⋅ √ a − √ a √ a − 1 ) = [ 1 + √ a ( √ a + 1 ) √ a + 1 ] [ 1 − √ a ( √ a − 1 ) √ a − 1 ] = ( 1 + √ a ) ( 1 − √ a ) = 1 − ( √ a ) 2 = 1 − a = V P
19 tháng 5 2021

a) VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}

=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}

=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}

=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}

=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}

=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4 \sqrt{6}}{2}\right) \cdot \dfrac{1}{\sqrt{6}}

=\left(\dfrac{-3}{2} \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}

=-\dfrac{3}{2}=-1,5=V P.
b) VT=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}

=\left(\dfrac{\sqrt{7} \cdot \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5} \cdot \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}

=\left[\dfrac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]: \dfrac{1}{\sqrt{7}-\sqrt{5}}

=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})

=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})

=-(7-5)=-2=VP.

c) V T=\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}

=\dfrac{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b}+\sqrt{b} \cdot \sqrt{b} \cdot \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}

=\dfrac{\sqrt{a} \cdot \sqrt{a b}+\sqrt{b} \cdot \sqrt{a b}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}

=\dfrac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})

=(\sqrt{a}+\sqrt{b}) \cdot(\sqrt{a}-\sqrt{b})

=a-b=V P.

d) VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)

=\left(1+\dfrac{\sqrt{a} \cdot \sqrt{a}+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a} \cdot \sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\right)

=\left[1+\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]

=(1+\sqrt{a})(1-\sqrt{a})

=1-(\sqrt{a})^{2}=1-a=V P

AH
Akai Haruma
Giáo viên
13 tháng 8 2021

j.

\(J=\left[\frac{1}{\sqrt{(\sqrt{5}-\sqrt{2})^2}}-\frac{\sqrt{2}}{\sqrt{2}(\sqrt{5}+\sqrt{2})}+1\right].\frac{1}{(\sqrt{2}+1)^2}\)

\(=\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right).\frac{1}{(\sqrt{2}+1)^2}\)

\(=[\frac{\sqrt{5}+\sqrt{2}-(\sqrt{5}-\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}+1].\frac{1}{(\sqrt{2}+1)^2}=(\frac{2\sqrt{2}}{3}+1).\frac{1}{(\sqrt{2}+1)^2}=\frac{3+2\sqrt{2}}{3}.\frac{1}{3+2\sqrt{2}}=\frac{1}{3}\)

AH
Akai Haruma
Giáo viên
13 tháng 8 2021

k. Đề sai sai, bạn xem lại

o.

\(O=(4+\sqrt{15})(\sqrt{5}-\sqrt{3}).\sqrt{2}.\sqrt{4-\sqrt{15}}\)

\(=(4+\sqrt{15}(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)

\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

 

20 tháng 8 2021

`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`

`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`

`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`

`=((9sqrt6)/6+(4sqrt6)/6)^2-24`

`=((13sqrt6)/6)^2-24`

`=13^2/6-24`

`=25/6`

23 tháng 5 2021

a) -17√3/3                                                  b) 11√6 

c) 21                                                            d) 11

29 tháng 5 2021

a)  a) Biến đổi vế trái thành 326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (6x+136x+6x):6x và làm tiếp

25 tháng 9 2021

a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)

b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)

d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)