Question

[Phạm Minh Thành]

Giải chi tiết ra giùm mk nhé. Cảm ơn

A> Cho biểu thức : \(A=\)\(\left(x^2-x-1\right)^2+2013\)

Tính giá trị của A khi x= \(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3+1}+1}}\)

B> cho \(\left(x+\sqrt{x^2+2013}\right)\cdot\left(y+\sqrt{y^2+2013}\right)=2013\). Chứng minh \(x^{2013}+y^{2013}=0\)

Answer 1

[Nguyễn Mai Quỳnh Anh]

B> \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(\left(x-\sqrt{x^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2013}=-x+\sqrt{x^2+2013}\)

Chứng minh tương tự: \(x+\sqrt{x^2+2013}=-y+\sqrt{y^2+2013}\)

cộng vế theo vế ta được: \(x+y=-x-y\)

\(\Leftrightarrow x+y=0\Leftrightarrow x=-y\Leftrightarrow x^{2013}=-y^{2013}\)

\(\Leftrightarrow x^{2013}+y^{2013}=0\)

Answer 2

[THIÊN SỨ LẠNH LÙNG]

a,Ta có x =...

x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3}+1}\right)\left(\sqrt{\sqrt{3}-1}\right)}\)

x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)

x = \(\frac{\sqrt{3}.2}{\sqrt{3}}\)

x = 2

sau đó thay x=2 vào A nhé.

A=2014 !!!