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15 tháng 7 2021

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

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

b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

15 tháng 7 2021

a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)

\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)

Vậy ta có đpcm 

b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)

Vậy ta có đpcm 

28 tháng 10 2021

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

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

24 tháng 10 2023

a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)

\(=1-x\sqrt{x}-x\sqrt{x}\)

\(=1-2x\sqrt{x}\)

b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)

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

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

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

28 tháng 10 2021

a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

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

13 tháng 11 2021

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

13 tháng 11 2021

a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)

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

8 tháng 5 2022

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

19 tháng 11 2021

\(a,VT=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{\left(x+1\right)\left(1-3x\right)}{3x}\right)\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2-6x}{3x}\right)\cdot\dfrac{x}{x-1}=\dfrac{6x}{3x}\cdot\dfrac{x}{x-1}=\dfrac{2}{x-1}=VP\left(x\ne0;x\ne1\right)\)

\(b,VT=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}=VP\left(a\ge0;a\ne1\right)\)

19 tháng 11 2021

anh Minh đâu r hả cj?

10 tháng 8 2018

1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn

10 tháng 8 2018

2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)

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