a) Ta có: \(\left\{{}\begin{matrix}\dfrac{2011}{2012}< \dfrac{2012}{2012}=1\\\dfrac{2012}{2011}>\dfrac{2011}{2011}=1\end{matrix}\right.\) => \(\dfrac{2011}{2012}< \dfrac{2012}{2011}\)

b) sửa đề: \(\dfrac{2010}{2013}\) và \(\dfrac{2011}{2012}\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{2010}{2013}< \dfrac{2010}{2012}\\\dfrac{2011}{2012}>\dfrac{2010}{20122}\end{matrix}\right.\) => \(\dfrac{2010}{2013}< \dfrac{2010}{2012}< \dfrac{2011}{2012}\)

=> \(\dfrac{2010}{2013}< \dfrac{2011}{2012}\)

P/s: Thật ra câu b ko cần sửa đề cx đc nhg mà thấy sai sai nên sửa thôi. Hjhj

a) \(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)

Đặt \(x^2+7=a\) . Thay vào PT ta được:

\(a+4x=\left(x+4\right)\sqrt{a}\)

<=> \(a+4x-x\sqrt{a}-4\sqrt{a}=0\)

<=> \(\sqrt{a}\left(\sqrt{a}-x\right)-4\left(\sqrt{a}-x\right)=0\)

<=> \(\left(\sqrt{a}-x\right)\left(\sqrt{a}-4\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{a}=x\\\sqrt{a}=4\end{matrix}\right.\) <=> \(\sqrt{a}=4\) ( Do \(\sqrt{a}=x\) vô nghiệm)

=> a = 16

=> \(x^2+7=16\) => \(x^2=9=>x=\pm3\)

Vậy nghiệm của PT: S = \(\left\{3;-3\right\}\)

P/s: Sai đừng trách nha!

Bài 2:

Ta có: \(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)

= \(\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

= \(a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)

= \(a^3+b^3+c^3+3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)

= \(a^3+b^3+c^3+3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)

= \(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\) => đpcm

Bài 1:

A = \(12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

=> \(\left(5^2-1\right)A\) = \(12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

=> 24A = \(12\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

=> A = \(\dfrac{12}{24}.\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

=> A = \(\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

=> A = \(\dfrac{1}{2}\left(5^{32}-1\right)\)

Ta có: \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)

= \(3n-2n^2-3+2n-n^2-5n\)

= \(-3n^2-3=3\left(1-n^2\right)\)

Vì 3 \(⋮\) 3 => 3 ( 1- n² ) \(⋮\) 3 => (n -1)( 3-2n) - n(n+5 ) \(⋮\) 3 với mọi x