\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)

\(2A=1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)

\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt:

\(B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)

\(2B=2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)

\(2B-B=\left(2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)

\(B=2-\dfrac{1}{2^{99}}\)

Vậy \(A=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}< 2\)

Bây giờ mình đang bận học bài 1 chút.Xíu nữa mình làm cho nhé

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)

\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)

Nên \(\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{y}{3}\end{matrix}\right.\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)

\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)

\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)

\(\circledast\) Với \(x=y=z\) thì \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

\(\circledast\) Với \(x+y+z=0\) thì\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

Khi đó \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)

\(M=\left|x+5\right|+\left|x-2\right|+\left(y-3\right)^2\)

\(M=\left|x+5\right|+\left|2-x\right|+\left(y-3\right)^2\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(M\ge\left|x+5+2-x\right|+\left(y-3\right)^2\)

\(M\ge7+\left(y-3\right)^2\)

Dấu "=" xảy ra khi:\(-5\le x\le2\)

Vì \(\left(y-3\right)^2\ge0\forall x\in R\) nên \(M\ge7+0=7\)

Dấu "=" xảy ra khi: \(y=3\)

Vậy \(min_M=7\) khi \(\left\{{}\begin{matrix}-5\le x\le2\\y=3\end{matrix}\right.\)

Giả sử điều cần chứng minh đúng thì:

\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

\(\Rightarrow\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1\ge\dfrac{3}{2}+1+1+1=\dfrac{9}{2}\)

\(\Rightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)

\(\Rightarrow\Rightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)

\(\Rightarrow\left(a+b+c+a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)

Đặt: \(\left\{{}\begin{matrix}b+c=x\\a+c=y\\a+b=z\end{matrix}\right.\) Khi đó:

\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)(đúng theo AM-GM)
Ta có đpcm

\(n^4+4\)

\(=n^4+4+4n^2-4n^2\)

\(=\left(n^2+2\right)^2-4n^2\)

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

Khi đó \(n^4+4\) là hợp số vì

\(n^4+4⋮\left\{{}\begin{matrix}n^2+2+2n\\n^2+2-2n\end{matrix}\right.\)