Prove that 1/2+√3 + 2/√5-√3 + 1/2-√5=0

Open in App

Solution

(1/2+√3)+(2/√5-√3)+(1/2-√5)
=1(2-√3)/(2-√3)(2+√3)+2(√5+√3)/(√5+√3)(√5-√3)+1(2+√5)/(2+√5)(2-√5)
=(2-√3)/{2²-(√3)²}+2(√5+√3)/{(√5)²-(√3)²}+(2+√5)/{2²-(√5)²}
=(2-√3)/(4-3)+2(√5+√3)/(5-3)+(2+√5)/(4-5)
=2-√3+{2(√5+√3)/2}-(2+√5)
=2-√3+√5+√3-2-√5
=0 (Proved)
(HOPE YOU LIKED MY ANSWER)

Suggest Corrections

228

Similar questions

x = 2^{1 / 3} + 2^{- 1 / 3}

2 x^{3} - 6 x - 5 = 0

(4, 5, - 5), (0, - 11, 3)

(2, - 3, - 1)

\int_{0}^{1} \frac{2 n + 3}{5 n^{2} + 1} d x

A = ⎛ ⎜ ⎝ \begin{matrix} 2 & 3 & 4 0 & - 2 & 1 3 & - 1 & 2 \end{matrix} ⎞ ⎟ ⎠, B = ⎛ ⎜ ⎝ \begin{matrix} 2 & 0 & - 3 4 & 0 & - 1 3 & 4 & 5 \end{matrix} ⎞ ⎟ ⎠

C = ⎛ ⎜ ⎝ \begin{matrix} 5 & 6 & 7 - 1 & 2 & 3 4 & - 5 & 4 \end{matrix} ⎞ ⎟ ⎠

A (B C)

(A B) C

Join BYJU'S Learning Program