haha Felix kìa :))

pt <=> \(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

<=> \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

<=> \(\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)

=> x + 2009 = 0 <=> x = -2009

Vậy phương trình có nghiệm x = -2009

Ta có : \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)(x;y \(\ne\)0)

<=> \(\frac{x+y}{xy}=\frac{1}{4}\)

<=> 4(x + y) = xy

<=> xy - 4x - 4y =0

<=> x(y - 4) - 4y + 16 = 16

<=> x(y - 4) - 4(y - 4) = 16

<=> (x - 4)(y - 4) = 16

Ta có 16 = 1.16 = 4.4 = (-4).(-4) = (-1).(-16)

Lập bảng xét các trường hợp

Vây các cặp (x;y) thỏa mãn là (5;20) ; (20;5) ; (8;8)

Áp dụng BĐT Bunyakovsky dạng phân thức ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\)

\(\ge\frac{\left(a+b+c\right)^2}{ab+bc+bc+ca+ca+ab}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c

a) Ta có: \(\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=\frac{7}{2}\)

\(\Leftrightarrow\left(8x+7\right)^2\cdot2\left(4x+3\right)\cdot8\left(x+1\right)=16\cdot\frac{7}{2}\)

\(\Leftrightarrow\left(8x+7\right)^2\left(8x+6\right)\left(8x+8\right)=56\)

Đặt \(8x+7=a\) khi đó:

\(a^2\left(a-1\right)\left(a+1\right)=56\)

\(\Leftrightarrow a^2\left(a^2-1\right)=56\)

\(\Leftrightarrow a^4-a^2-56=0\)

\(\Leftrightarrow\left(a^2-8\right)\left(a^2+7\right)=0\)

\(\Leftrightarrow a^2-8=0\Leftrightarrow\left(8x+7\right)^2-8=0\)

\(\Leftrightarrow\left(8x+7\right)^2=8\Leftrightarrow\orbr{\begin{cases}8x+7=2\sqrt{2}\\8x+7=-2\sqrt{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}8x=2\sqrt{2}-7\\8x=-2\sqrt{2}-7\end{cases}}\Rightarrow x=\frac{\pm2\sqrt{2}-7}{8}\)

b) Ta có: \(x^2+5y^2-4xy+10x-22y+\left|x+y+z\right|+26=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+25+y^2-2y+1+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y-1\right)^2+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\\z=2\end{cases}}\)

Vậy x = -3 , y = 1 , z = 2

Gọi \(d=\left(n^3+2n;n^4+3n^2+1\right)\)

\(\Rightarrow\hept{\begin{cases}\left(n^3+2n\right)⋮d\\\left(n^4+3n^2+1\right)⋮d\end{cases}}\Leftrightarrow\hept{\begin{cases}n\left(n^3+2n\right)=\left(n^4+2n^2\right)⋮d\\\left(n^4+3n^2+1\right)⋮d\end{cases}}\)

\(\Rightarrow\left(n^4+3n^2+1\right)-\left(n^4+2n^2\right)⋮d\)

\(\Leftrightarrow n^2+1⋮d\Leftrightarrow\left(n^2+1\right)^2⋮d\)

\(\Rightarrow\left(n^2+1\right)^2-\left(n^4+2n^2\right)⋮d\Leftrightarrow1⋮d\Rightarrow d=1\)

=> P/s tối giản

Gọi \(d=ƯCLN\left(n^3+2n;n^4+3n^2+1\right);\left(d>0\right)\)

\(\Rightarrow\hept{\begin{cases}n^3+2n⋮d\left(1\right)\\n^4+3n^2+1⋮d\end{cases}}\)

Từ \(\left(1\right)\): \(\Rightarrow n\left(n^3+2n\right)⋮d\)

\(\Rightarrow n^4+2n^2⋮d\)

\(\Rightarrow\left(n^4+3n^2+1\right)-\left(n^4+2n^2\right)⋮d\)

\(\Rightarrow n^2+1⋮d\)

\(\Rightarrow\left(n^2+1\right)^2⋮d\)

\(\Rightarrow n^4+2n^2+1⋮d\)

\(\Rightarrow1⋮d\)(do \(n^4+2n^2⋮d\))

Vì \(d>0\)\(\Rightarrow d=1\)

\(\Rightarrow\left(n^3+2n;n^4+3n^2+1\right)=1\)

\(\Rightarrow\frac{n^3+2n}{n^4+3n^2+1}\)là phân số tối tối giản với mọi n nguyên

Ta có: \(P=\frac{2016x^2-2x+1}{x^2}=\frac{2015x^2+\left(x^2-2x+1\right)}{x^2}\)

\(=2015+\frac{\left(x-1\right)^2}{x^2}\ge2015\left(\forall x\ne0\right)\)

Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy Min(P) = 2015 khi x = 1

Ta có : \(P=\frac{2016x^2-2x+1}{x^2}\)

\(=\frac{2015x^2+\left(x-1\right)^2}{x^2}\)

\(=2015+\left(\frac{x-1}{x}\right)^2\)

Vì \(\left(\frac{x-1}{x}\right)^2\ge0\forall x\ne0\)

\(\Rightarrow P\ge2015\forall x\ne0\)

Dấu \("="\) xảy ra \(\Leftrightarrow\left(\frac{x-1}{x}\right)^2=0\)

\(\Leftrightarrow\frac{x-1}{x}=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy \(MinP=2015\Leftrightarrow x=1\)

Từ \(\hept{\begin{cases}2a=by+cz\\2b=cz+ax\\2c=ax+by\end{cases}}\)

\(\Rightarrow2\left(a+b+c\right)=2\left(ax+by+cz\right)\)

\(\Rightarrow a+b+c=ax+by+cz=ax+2a=a\left(x+2\right)\)

\(\Rightarrow\frac{1}{x+2}=\frac{a}{a+b+c}\)

Tương tự , ta có : \(\hept{\begin{cases}\frac{1}{y+2}=\frac{b}{a+b+c}\\\frac{1}{z+2}=\frac{c}{a+b+c}\end{cases}}\)

\(\Rightarrow A=\frac{a+b+c}{a+b+c}=1\)

Vậy \(A=1\)

mình ghi thiếu: VP=\(\frac{1}{9}\)

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{1}{9}\)

<=> \(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{1}{9}\)

<=> \(\frac{1}{2}\left(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}\right)=\frac{1}{9}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}=\frac{2}{9}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{9}\)

<=> \(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\)

<=> (x + 1)(x + 7) = 27

<=> x² + 8x + 7 - 27 = 0

<=> x² + 8x - 20 = 0

<=> x² - 2x + 10x - 20 = 0

<=> x(x - 2) + 10(x - 2) = 0

<=> (x + 10(x - 2) = 0

<=> \(\orbr{\begin{cases}x=-10\\x=2\end{cases}}\) 

Vậy \(x\in\left\{-10;2\right\}\)là giá trị cần tìm