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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: \(D=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\dfrac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a.\sqrt{a}.\left(\sqrt{b}-\sqrt{a}\right)+b.\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(b-a\right)}{\sqrt{ab}\left(b-a\right)}\right)\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a\sqrt{ab}-a^2+b\sqrt{ab}+b^2-b^2+a^2}{\sqrt{ab}\left(b-a\right)}\right)\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a\sqrt{ab}+b\sqrt{ab}}{\sqrt{ab}\left(b-a\right)}\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(b-a\right)}=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a+b}{b-a}\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}.\dfrac{b-a}{a+b}=\dfrac{b-a}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{a}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{b}-\sqrt{a}\)
\(D=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a^2-b^2\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{a^2-a\sqrt{ab}-b\sqrt{ab}-b^2-a^2+b^2}{\sqrt{ab}\left(a-b\right)}\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{-\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\dfrac{a+b+b\sqrt{a}-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{-\left(a-b\right)}{a+b}\)
\(=\dfrac{-\left(a+b+b\sqrt{a}-\sqrt{ab}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)
Mạn phép ko chép lại đề , mk làm luôn
a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)
\(D=\dfrac{2\left(\sqrt{a}+\sqrt{b}\right)}{\left(b+1\right)\left(a+1\right)}\)
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
\(a)A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\sqrt{5}+3+\sqrt{3}-\left(\sqrt{5}+3\right)\\ =\sqrt{3}\)
\(b)B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2\\ =\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\\ =2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\\ =2\left(25-21\right)=8\)
ĐKXĐ: \(a,b\ge0,a\ne b\)
= \(\dfrac{3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}-b\sqrt{b}+2a\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)
= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\)
= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
= \(\dfrac{3\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=3\)