Ta có: \(A=\left|2x-2\right|+\left|2013-2x\right|\ge\left|2x-2+2013-2x\right|=2011\)

Dấu '=' xảy ra khi \(\left(2x-2\right)\left(2013-2x\right)\ge0\Leftrightarrow1\le x\le\frac{2013}{2}\)

Ta có:

\(|x|\ge0\)với mọi \(x\)

\(|x-2|\ge0\)với mọi \(x\)

Do đó:

\(A=x-\left(x-2\right)\)

\(A=x-x+2\)

\(A=0+2\)

\(A=2\)

TH1 \(x< 0\)   

\(A=|x|-|x-2|\)   

\(=\left(-x\right)-\left[-\left(x-2\right)\right]\)   

\(=-x-\left(2-x\right)\)

\(=-x+x-2\)   

\(=-2\)   

TH2 \(0\le x< 2\)   

\(A=|x|-|x-2|\)   

\(=x-\left[-\left(x-2\right)\right]\)

\(=x-\left(2-x\right)\)   

\(=2x-2\)   

TH3 \(x\ge2\)   

\(A=|x|-|x-2|\)    

\(=x-\left(x-2\right)\)   

\(=2\)

\(\left(\frac{3}{4}\right)^x=\frac{2^8}{3^4}\Leftrightarrow\frac{3^x}{4^x}=\frac{2^8}{3^4}\)

\(\Leftrightarrow3^x.3^4=2^8.4^x\Leftrightarrow3^{x+4}=\left(2^2\right)^4.4^x\)

\(\Leftrightarrow3^{x+4}=4^{4+x}\Leftrightarrow3^{x+4}-4^{x+4}=0\)xD 

a) P(-1)=(-1)^2+.................+(-1)^114

P(-1)=1+.....................+1

P(-1)=57

b)Q(-1)=-1+....................+(-1)^115

Q(-1)=-1+..................+(-1)

Q(-1)=-58

\(1\frac{3x}{4}+1\frac{1}{2}=-\frac{4}{5}\)

\(\Leftrightarrow\frac{7x}{4}+\frac{3}{2}=-\frac{4}{5}\Leftrightarrow\frac{35x}{20}+\frac{30}{20}=-\frac{16}{20}\)

\(\Rightarrow35x=-46\Leftrightarrow x=-\frac{46}{35}\)

Tuyển gái để chat sex, và địt tưởng tượng

\(xy=\frac{1}{t}.txy\le\frac{t^2x^2+y^2}{2t}=\frac{\left(3+\sqrt{5}\right)x^2+y^2}{1+\sqrt{5}}\)\(t^2=\frac{3+\sqrt{5}}{2}\)

\(\frac{2\left(1+\sqrt{5}\right)\left(x^2+y^2+z^2+1\right)}{\left(3+\sqrt{5}\right)\left(2x^2+y^2+z^2+1\right)}\)

\(K=\frac{x^2+y^2+z^2+1}{xy+yz+z}=\frac{\left(1+\sqrt{5}\right)\left(x^2+y^2+z^2+1\right)}{2.\frac{1+\sqrt{5}}{2}x.y+\left(1+\sqrt{5}\right)yz+2.\frac{1+\sqrt{5}}{2}.z}\)

\(\ge\frac{\left(1+\sqrt{5}\right)\left(x^2+y^2+z^2+1\right)}{\frac{3+\sqrt{5}}{2}x^2+y^2+\frac{1+\sqrt{5}}{2}\left(y^2+z^2\right)+z^2+\frac{3+\sqrt{5}}{2}}=\frac{1+\sqrt{5}}{\frac{3+\sqrt{5}}{2}}=\sqrt{5}-1=k\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=1\\y=\frac{1+\sqrt{5}}{2}\\z=\frac{1+\sqrt{5}}{2}\end{cases}}\)

\(M=\frac{x^2+y^2+z^2+1}{xy+y+z}=\frac{\left(\sqrt{5}-1\right)\left(x^2+y^2+z^2+1\right)}{2.x.\frac{\sqrt{5}-1}{2}y+\left(\sqrt{5}-1\right)y+2.\frac{\sqrt{5}-1}{2}.z}\)

\(\ge\frac{\left(\sqrt{5}-1\right)\left(x^2+y^2+z^2+1\right)}{x^2+\frac{3-\sqrt{5}}{2}y^2+\frac{\sqrt{5}-1}{2}\left(y^2+1\right)+\frac{3-\sqrt{5}}{2}+z^2}=\sqrt{5}-1=m\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\\y=1\\z=\frac{-1+\sqrt{5}}{2}\end{cases}}\)

\(km+k+m=4\)

2 dòng đầu sai nhưng quên xoá :) bỏ đi nhé 

Theo tính chất dãy tỉ số bằng nhau ta có:

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

\(\Rightarrow\frac{2a+b}{c}=\frac{3}{3}=1=\frac{a}{2b+c}=\frac{3b}{2c+a}\)

Vậy \(\frac{2a+b}{c}=\frac{a}{2b+c}=\frac{3b}{2c+a}=1\)

vậy nếu a+b+c = 0 thì sao ?