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20 tháng 5 2017

a , ĐKXĐ : bạn tự làm nhé

Đặt \(\sqrt{a}=x\) , khi đó biểu thức trỏ thành:

P = \(\dfrac{x^2.x-1}{x^2-x}-\dfrac{x^2.x+1}{x^2+x}+\left(x-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right)\)

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

= \(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)x}-\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)x}+\dfrac{\left(x^2+2x+1\right)+\left(x^2-2x+1\right)}{x}\)

= \(\dfrac{x^2+x+1+x^2-x+1}{x}+\dfrac{2x^2+2}{x}\)

= \(\dfrac{4x^2+4}{x}\)= \(\dfrac{4a+4}{\sqrt{a}}\)

Mấy câu sau dễ rồi

21 tháng 5 2017

xie xie! cảm ơn bạn nha!

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

b:Sửa đề: 2A

2A=2căn x+5

=>(2căn x+2)/căn x=2căn x+5

=>2x+5căn x-2căn x-2=0

=>2x+3căn x-2=0

=>(căn x+2)(2căn x-1)=0

=>x=1/4

16 tháng 11 2021

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

23 tháng 8 2017

\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)

\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

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

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

13 tháng 6 2018

Mạn phép ko chép lại đề , mk làm luôn

a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)

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

13 tháng 6 2018

D=A/B

a)

B=1+(a+b+2ab)/(1-ab)=(a+b+ab)/(1-ab)

dk: a,b≥0; a.b≠1

1/B=(1-ab)/(a+b+ab)

A=√a+√b)[(1+√ab)+(1-√ab)]/(1-ab)=2(√a+√b)/(1-ab)

D=2(√a+√b)/[(a+1)(b+1)]

b)

a=2/(√3+2)=2(2-√3)/[(2+√3)(2-√3)]=2(2-√3)=(√3-1)^2

Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)

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

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

\(A=\left[1:\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right)\right]\cdot\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)

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

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

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