Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Minh Phương Đặng
Xem chi tiết
Thái Xuân Đăng
26 tháng 12 2015 lúc 19:39

\(P=\left(\frac{3\sqrt{a}}{a+\sqrt{ab}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)

\(=\left(\frac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{3a}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\left(a+\sqrt{ab}+b\right)}{\left(a+\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right).\frac{2\left(a+\sqrt{ab}+b\right)}{\cdot\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

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

\(=\frac{2}{a-1}\)

nguyen thu phuong
Xem chi tiết
Nguyễn Nhã Thanh
Xem chi tiết
Khang Võ
Xem chi tiết
PHAM THANH THUONG
Xem chi tiết
Pham Thanh Thuong
Xem chi tiết
Nguyễn Nhã Thanh
Xem chi tiết
Trình
20 tháng 8 2017 lúc 16:42

\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)

\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)

\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)

\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)


Điều kiện : a, b\(\ge0\)

Crackinh
Xem chi tiết
Lê Ng Hải Anh
29 tháng 8 2018 lúc 22:00

Ta có:

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

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

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

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

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

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

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

\(=\frac{3a-3b}{a-b}\)

\(=3\)

=.= hok tốt!!

Lê Lan Hương
Xem chi tiết