.......
Bila f(x) =![\frac{x}{a} [ \frac{b^2}{x^2} ] + \frac{x}{b} [1- \frac{a^2}{x^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Ba%7D+%5B+%5Cfrac%7Bb%5E2%7D%7Bx%5E2%7D+%5D+%2B+%5Cfrac%7Bx%7D%7Bb%7D+%5B1-+%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D+%5D "\frac{x}{a} [ \frac{b^2}{x^2} ] + \frac{x}{b} [1- \frac{a^2}{x^2} ]")
, maka f(a+b)= ...
Bila f(x) =
ekamiaww
![f(x)= \frac{x}{a} [ \frac{b^2}{x^2}]+\frac{x}{b} [1- \frac{a^2}{x^2}] \\ f(a+b)=\frac{a+b}{a} [ \frac{b^2}{(a+b)^2}]+\frac{a+b}{b} [1- \frac{a^2}{(a+b)^2}] \\ =\frac{a+b}{a} . \frac{b^2}{(a+b)^2}+\frac{a+b}{b}-\frac{a+b}{b}. \frac{a^2}{(a+b)^2} \\ = \frac{b^2}{a(a+b)} +\frac{a+b}{b}- \frac{a^2}{b(a+b)} \\ = \frac{b^3+a(a+b)(a+b)-a^3}{ab(a+b)} \\ = \frac{b^3+a(a^2+2ab+b^2)-a^3}{ab(a+b)} \\ =\frac{b^3+a^3+2a^2b+ab^2-a^3}{ab(a+b)} \\ =\frac{b^3+2a^2b+ab^2}{ab(a+b)}](https://tex.z-dn.net/?f=f%28x%29%3D+%5Cfrac%7Bx%7D%7Ba%7D+%5B+%5Cfrac%7Bb%5E2%7D%7Bx%5E2%7D%5D%2B%5Cfrac%7Bx%7D%7Bb%7D+%5B1-+%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5D++%5C%5C+f%28a%2Bb%29%3D%5Cfrac%7Ba%2Bb%7D%7Ba%7D+%5B+%5Cfrac%7Bb%5E2%7D%7B%28a%2Bb%29%5E2%7D%5D%2B%5Cfrac%7Ba%2Bb%7D%7Bb%7D+%5B1-+%5Cfrac%7Ba%5E2%7D%7B%28a%2Bb%29%5E2%7D%5D++%5C%5C+%3D%5Cfrac%7Ba%2Bb%7D%7Ba%7D+.+%5Cfrac%7Bb%5E2%7D%7B%28a%2Bb%29%5E2%7D%2B%5Cfrac%7Ba%2Bb%7D%7Bb%7D-%5Cfrac%7Ba%2Bb%7D%7Bb%7D.+%5Cfrac%7Ba%5E2%7D%7B%28a%2Bb%29%5E2%7D+%5C%5C+%3D+%5Cfrac%7Bb%5E2%7D%7Ba%28a%2Bb%29%7D+%2B%5Cfrac%7Ba%2Bb%7D%7Bb%7D-+%5Cfrac%7Ba%5E2%7D%7Bb%28a%2Bb%29%7D++%5C%5C+%3D+%5Cfrac%7Bb%5E3%2Ba%28a%2Bb%29%28a%2Bb%29-a%5E3%7D%7Bab%28a%2Bb%29%7D++%5C%5C+%3D+%5Cfrac%7Bb%5E3%2Ba%28a%5E2%2B2ab%2Bb%5E2%29-a%5E3%7D%7Bab%28a%2Bb%29%7D++%5C%5C+%3D%5Cfrac%7Bb%5E3%2Ba%5E3%2B2a%5E2b%2Bab%5E2-a%5E3%7D%7Bab%28a%2Bb%29%7D++%5C%5C+%3D%5Cfrac%7Bb%5E3%2B2a%5E2b%2Bab%5E2%7D%7Bab%28a%2Bb%29%7D+ "f(x)= \frac{x}{a} [ \frac{b^2}{x^2}]+\frac{x}{b} [1- \frac{a^2}{x^2}] \\ f(a+b)=\frac{a+b}{a} [ \frac{b^2}{(a+b)^2}]+\frac{a+b}{b} [1- \frac{a^2}{(a+b)^2}] \\ =\frac{a+b}{a} . \frac{b^2}{(a+b)^2}+\frac{a+b}{b}-\frac{a+b}{b}. \frac{a^2}{(a+b)^2} \\ = \frac{b^2}{a(a+b)} +\frac{a+b}{b}- \frac{a^2}{b(a+b)} \\ = \frac{b^3+a(a+b)(a+b)-a^3}{ab(a+b)} \\ = \frac{b^3+a(a^2+2ab+b^2)-a^3}{ab(a+b)} \\ =\frac{b^3+a^3+2a^2b+ab^2-a^3}{ab(a+b)} \\ =\frac{b^3+2a^2b+ab^2}{ab(a+b)}")
1 votes
Thanks 1
More Questions From This User See All