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\(D=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)

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

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

\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{-\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}\)

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

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

10 tháng 6 2017

Bài này chính xác là của lớp 9 nè!!

Đề bạn ghi sai hay sao ý, pn xem lại xem, mk sửa đề như dưới, pn tham khảo:

Ta có: \(D=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)

\(=\dfrac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a.\sqrt{a}.\left(\sqrt{b}-\sqrt{a}\right)+b.\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(b-a\right)}{\sqrt{ab}\left(b-a\right)}\right)\)

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

\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a\sqrt{ab}+b\sqrt{ab}}{\sqrt{ab}\left(b-a\right)}\)

\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(b-a\right)}=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a+b}{b-a}\)

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

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

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

17 tháng 5 2017

ĐKXĐ: \(a,b\ge0,a\ne b\)

= \(\dfrac{3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}-b\sqrt{b}+2a\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)

= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)

= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)

= \(\dfrac{3\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=3\)

16 tháng 11 2021

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