\(A^2=2012^2+2012^2.2013^2+2013^2\)

\(A^2=\left(2013-1\right)^2+2013^2+2012^2.2013^2\)

\(A^2=2.2013^2-2.2013+1+2012^2.2013^2\)

\(A^2=2012^2.2013^2+2.2013.\left(2013-1\right)+1\)

\(A^2=\left(2012.2013+1\right)^2\Rightarrow A=2012.2013+1\) là số tự nhiên

a.

\(a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2=a^2+\left(a^2+a\right)^2+a^2+2a+1\)

\(=\left(a^2+a\right)^2+2\left(a^2+a\right)+1=\left(a^2+a+1\right)^2\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=3\\x+\dfrac{1}{y}+\dfrac{x}{y}=3\end{matrix}\right.\)

\(\Rightarrow\left(x+\dfrac{1}{y}\right)^2+x+\dfrac{1}{y}=6\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=2\Rightarrow\dfrac{x}{y}=1\\x+\dfrac{1}{y}=-3\Rightarrow\dfrac{x}{y}=6\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=2\\\dfrac{x}{y}=1\end{matrix}\right.\) \(\Rightarrow...\)

Cho hình bình hành ABCD,cạnh AB=a.AD=b .Tính AC^2+BD^2 theo a và b

giúp em với ạ

\(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\Rightarrow A^2=2012^2+2012^2.2013^2+2013^2=2012^2.2013^2+\left(2013-1\right)^2+2013^2=\left(2012.2013\right)^2+2013^2-2.2013+1+2013=\left(2012.2013\right)^2+2.2013^2-2.2013+1=\left(2012.2013\right)^2+2.2013\left(2013-1\right)+1=\left(2012.2013\right)^2+2.2012.2013.1+1=\left(2012.2013+1\right)^2\Rightarrow A=2012.2013+1\)Vậy A là một số tự nhiên

A²=2012²+2012²2013²+2013²

A²=(2013-1)²+2013²+2012²2013²

A²=2.2013²-2.2013+1+2012²2013²

A²=2012²2013²+2.2013(2013-1)+1

A²=(2012.2013+1)² \(\Rightarrow\)A=2012.2013+1 la so tu nhien

ĐKXĐ: \(x-2013\ge0\Leftrightarrow x\ge2013\)

Ta có:

\(A=\sqrt{x-2013-2\sqrt{x-2013}+1}+\sqrt{x-2013-90\sqrt{x-2013}+2025}\)

\(=\sqrt{\left(\sqrt{x-2013}-1\right)^2}+\sqrt{\left(\sqrt{x-2013}-45\right)^2}\)

\(=\left|\sqrt{x-2013}-1\right|+\left|\sqrt{x-2013}-45\right|\)

\(=\left|\sqrt{x-2013}-1\right|+\left|45-\sqrt{x-2013}\right|\)

\(\ge\left|\sqrt{x-2013}-1+45-\sqrt{x-2013}\right|\)

\(=\left|-1+45\right|=\left|44\right|=44\)

Vậy GTNN của A là 44, đạt được khi và chỉ khi \(\left(\sqrt{x-2013}-1\right)\left(45-\sqrt{x-2013}\right)\ge0\)

\(\Leftrightarrow1\le\sqrt{x-2013}\le45\)

\(\Leftrightarrow1\le x-2013\le2025\)

\(\Leftrightarrow2014\le x\le4038\left(tm\right)\)

Lời giải:

$A=|1-\sqrt{2012}|\sqrt{2012+2\sqrt{2012}+1}$

$=|1-\sqrt{2012}|\sqrt{(\sqrt{2012}+1)^2}$

$=|1-\sqrt{2012}|.|\sqrt{2012}+1|$

$=|(1-\sqrt{2012})(1+\sqrt{2012})|=|1-2012|=2011$

Đặt t= 2012

Thay vào ta được :\(\sqrt{t^2+t^2\left(t+1\right)^2+\left(t+1\right)^2}=\sqrt{t^2+t^4+2t^3+t^2+t^2+2t+1}\)

                            =\(\sqrt{t^4+t^2+1+2\left(t^3+t^2+t\right)}=\sqrt{\left(t^2+t+1\right)^2}=t^2+t+1\)

                            = \(2012^2+2012+1\)là số tự nhiên (đpcm)

k biết lm thì nói toạt ra lun đi

minh bik lam ne

đặt a =2012

\(\Rightarrow A=\sqrt{a^2+a^2\left[a+1\right]^2+\left\{a+1\right\}^2}\)

           \(=\sqrt{a^2+a^4+2a^3+a^2+2a+1}\)

           \(=\sqrt{a^4+2a^3+3a^2+2a+1}\)

           \(=\sqrt{\left[a^2+a+1\right]^2}\)

           \(=a^2+a+1\)

           \(=2012^2+2012+1\) là 1 số tự nhiên