Chắc là a;b;c hết chứ?

\(VT=\dfrac{a}{a+b+c+b-a}+\dfrac{b}{a+b+c+c-b}+\dfrac{c}{a+b+c+a-c}\)

\(VT=\dfrac{a}{c+2b}+\dfrac{b}{a+2c}+\dfrac{c}{b+2a}=\dfrac{a^2}{ac+2ab}+\dfrac{b^2}{ab+2bc}+\dfrac{c^2}{bc+2ac}\)

\(VT\ge\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\) (đpcm)

cho x,y,z>0 ,x+y+z=1 chu nhi?

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

\(\Rightarrow\dfrac{y}{1+z-y}=\dfrac{y}{x+y+z+z-y}=\dfrac{y}{2z+x}\)

\(\Rightarrow\dfrac{z}{1+x-z}=\dfrac{z}{x+y+z+x-z}=\dfrac{z}{2x+y}\)

\(\Rightarrow A=\dfrac{x}{2y+z}+\dfrac{y}{2z+x}+\dfrac{z}{2x+y}=\dfrac{x^2}{2xy+xz}+\dfrac{y^2}{2zy+xy}+\dfrac{z^2}{2xz+xz}\ge\dfrac{\left(x+y+z\right)^2}{3\left(xy+yz+xz\right)}=1\)

dau"=" xay ra<=>x=y=z=1/3

\(B=\frac{x+5}{2x}+\frac{6-x}{x-5}+\frac{2x^2-9x+25}{2x\left(x-5\right)}\)

\(=\frac{\left(x+5\right)\left(x-5\right)+2x\left(6-x\right)+2x^2-9x+25}{2x\left(x-5\right)}\)

\(=\frac{x^2-25+12x-2x^2+2x^2-9x+25}{2x\left(x-5\right)}=\frac{x^2+3x}{2x\left(x-5\right)}\)

\(=\frac{x\left(x+3\right)}{2x\cdot\left(x-5\right)}=\frac{x+3}{2\left(x-5\right)}\)

\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)

\(=abc-\left(ab+bc+ca\right)+a+b+c-1\)

\(=abc-abc+1-1=0\) (đpcm)

Ta có x⁰= 1 ( x khác 0)

Vì nó có định lý trong sgk toán

vì x⁰=1

mà theo tính chất x⁰=1

=> x\(\in\)N*

Với mọi giá trị x mà x⁰ thì luôn bằng 

Ví dụ: 5⁰ = 1; 26⁰ = 1; ........