Processing math: 100%

Thursday, October 14, 2021

2021/084) a,b,c are positive real numbers such that a^2+b^2= c^2 and ab=c find the value of \frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}

 We have

a^2+b^=c^2\cdots(1)

And ab = c\cdots(2)

Now

\frac{(a+b+c)(a+b-c)}{c} = \frac{(a+b^2) - c^2 }{c} = \frac{a^2+b^2+2ab-c^2}{c} = \frac{2ab}{c} (uing (1)

= \frac{2c}{c} (sing (2)

= 2

Or

\frac{(a+b+c)(a+b-c)}{c^2} = 2

similarly

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

multiplying we get \frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}= 4


No comments: