# Ngô Quốc Anh

## September 13, 2009

### An algebraic identity for 9th level

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 21:16

In this topic I will show you how to prove $\sqrt{2} + \sqrt{{20}} - \sqrt{{25}} = 3\sqrt {\sqrt{5} - \sqrt{4}}$

In order to prove that fact, we just do as the following: by using $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$

we obtain ${\Big( {\sqrt{2} + \underbrace {\sqrt{{20}}}_{\sqrt{{{2^2}}}\sqrt{5}} - \underbrace {\sqrt{{25}}}_{\sqrt{{{5^2}}}}} \Big)^2}=\sqrt{{{2^2}}} + \sqrt{{{2^4}}}\sqrt{{{5^2}}} + \sqrt{{{5^4}}} + 2\left( {\underbrace {\sqrt{2}\sqrt{{{2^2}}}}_2\sqrt{5} - \sqrt{2}\sqrt{{{5^2}}} - \sqrt{{{2^2}}}\underbrace {\sqrt{5}\sqrt{{{5^2}}}}_5} \right).$

Therefore ${\left( {\sqrt{2} + \sqrt{{20}} - \sqrt{{25}}} \right)^2} = \sqrt{{{2^2}}} + 2\sqrt{2}\sqrt{{{5^2}}} + 5\sqrt{5} + 2\left( {2\sqrt{5} - \sqrt{2}\sqrt{{{5^2}}} - 5\sqrt{{{2^2}}}} \right).$

Clearly $\sqrt{{{2^2}}} + 2\sqrt{2}\sqrt{{{5^2}}} + 5\sqrt{5} + 2\left( {2\sqrt{5} - \sqrt{2}\sqrt{{{5^2}}} - 5\sqrt{{{2^2}}}} \right) = 9\left( {\sqrt{5} - \sqrt{{{2^2}}}} \right).$

Thus ${\left( {\sqrt{2} + \sqrt{{20}} - \sqrt{{25}}} \right)^2} = 9\left( {\sqrt{5} - \sqrt{4}} \right)$

which yields $\left| {\sqrt{2} + \sqrt{{20}} - \sqrt{{25}}} \right| = 3\sqrt {\sqrt{5} - \sqrt{4}} .$

Finally, by using the fact that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ we get ${\left( {\sqrt{2} + \sqrt{{20}}} \right)^3} - 25 = \underbrace {\left( {22 + 3\sqrt{{{2^2}}}\sqrt{{20}} + 3\sqrt{2}\sqrt{{{{20}^2}}}} \right) - 25}_{3\left( {\sqrt{{{2^2}}}\sqrt{{20}} + \sqrt{2}\sqrt{{{{20}^2}}} - 1} \right)} > 0$

which implies ${\sqrt{2} + \sqrt{{20}} - \sqrt{{25}}}>0.$

In other words, $\sqrt{2} + \sqrt{{20}} - \sqrt{{25}} = 3\sqrt {\sqrt{5} - \sqrt{4}}.$

## 3 Comments »

1. Hi !
Liệu đẳng thức sau có bằng nhau không ?

Comment by tamddao — December 16, 2009 @ 9:34

2.  $\sqrt{2}+\sqrt{20}-\sqrt{25}=\sqrt{-63-9\sqrt{10}+18\sqrt{100}}$

Comment by tamddao — December 16, 2009 @ 10:30

• Motivation của nó là gì?

Comment by Ngô Quốc Anh — December 16, 2009 @ 11:30

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