Question

Thực hiện phép tính :

a)        \(\left(\frac{x-y}{x+y}+\frac{x+y}{x-y}\right).\left(\frac{x^2+y^2}{2xy}+1\right).\frac{xy}{x^2+y^2}\)

b)             \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

P/s : Giúp chauu với nhá mấy bacc kewtt :>>

Answer 1

\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Answer 2

Sr nha

Kq mik nhầm

Ko phải -2b đâu mà = 0

Answer 3

Oce :) Thks Moon :33

Answer 4

a) \(\left(\frac{x-y}{x+y}+\frac{x+y}{x-y}\right)\left(\frac{x^2+y^2}{2xy}+1\right)\cdot\frac{xy}{x^2+y^2}\)

= \(\left[\frac{\left(x-y\right)^2}{\left(x+y\right)\left(x-y\right)}+\frac{\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)}\right]\left(\frac{x^2+y^2+2xy}{2xy}\right)\cdot\frac{xy}{x^2+y^2}\)

= \(\left[\frac{x^2-2xy+y^2+x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\right]\left[\frac{\left(x+y\right)^2}{2xy}\right]\cdot\frac{xy}{x^2+y^2}\)

= \(\frac{2x^2+2y^2}{\left(x+y\right)\left(x-y\right)}\cdot\frac{\left(x+y\right)^2}{2xy}\cdot\frac{xy}{x^2+y^2}\)

= \(\frac{2\left(x^2+y^2\right)\left(x+y\right)^2.xy}{\left(x+y\right)\left(x-y\right)\cdot2xy\cdot\left(x^2+y^2\right)}\)

= \(\frac{x+y}{x-y}\)

Answer 5

Thks Edogawa Conan nhèo thiệt nhèo :33