\(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}=\frac{a^2+b^2+c^2}{abc}\ge\frac{ab+bc+ca}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Vì \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Bài 1:

\(\sqrt{1}-\sqrt{4}+\sqrt{9}-\sqrt{16}+\sqrt{25}-\sqrt{36}+.....-\sqrt{400}\)

\(=1-2+3-4+5-6+.....-20\)

\(=\left(1-2\right)+\left(3-4\right)-\left(5-6\right)+.....+\left(19-20\right)\)

\(=\left(-1\right)\times\dfrac{\dfrac{\left(20-1\right)\times1+1}{2}}{2}\)

\(=\left(-1\right)\times10\)

\(=-10\)

Dễ thế này mà ko ai lm à

Chúc bn học tốt

\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

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

\(=3^n\times10-2^n\times5\)

\(=3^n\times10-2^{n-1}\times2\times5\)

\(=3^n\times10-2^{n-1}\times10\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

Đến đây bn kết nốt

Chúc bn học tốt

\(7\times3^{x-1}-3^{x+2}=-540\)

\(7\times3^{x-1}-3^{x-1}\times3^3=-540\)

\(3^{x-1}\left(7-3^3\right)=-540\)

\(3^{x-1}\left(7-27\right)=-540\)

\(3^{x-1}\times\left(-20\right)=-540\)

\(3^{x-1}=\left(-540\right)\div\left(-20\right)\)

\(3^{x-1}=27\)

\(3^{x-1}=3^3\)

\(\Rightarrow x-1=3\)

\(x=4\)

Chúc bn học tốt

1. Gọi 4 số nguyên liên tiếp đó là \(x;x+1;x+2;x+3\left(x\in Z\right)\)

Ta có \(\left(x+1\right)\left(x+3\right)-x\left(x+2\right)=99\)

\(\left(x+1\right)x+\left(x+1\right)3-x^2-2x=99\)

\(x^2+x+3x+3-x^2-2x=99\)

\(\left(x^2-x^2\right)+\left(x+3x-2x\right)+3=99\)

\(2x=99-3\)

\(2x=96\)

\(x=48\)

\(\Rightarrow x+1=49\)

\(x+2=50\)

\(x+3=51\)

Vậy 4 số nguyên đó là 48;49;50;51

Chúc bn học tốt

\(11\times121^{1006}\le11^n\le11^{2015}\)

\(11\times\left(11^2\right)^{1006}\le11^n\le11^{2015}\)

\(11\times11^{2012}\le11^n\le11^{2015}\)

\(11^{2013}\le11^n\le11^{2015}\)

\(\Rightarrow2013\le n\le2015\)

\(\Rightarrow n\in\left\{2013;2014;2015\right\}\)

Chúc bn học tốt