Cho a,b,c > 0 và ab + bc + ca = 3. Chứng minh rằng :
\(\sqrt[3]{1+3a^2}+\sqrt[3]{1+3b^2}+\sqrt[3]{1+3c^2}\ge2\left(\sqrt[3]{\frac{a}{b+c}}+\sqrt[3]{\frac{b}{c+a}}+\sqrt[3]{\frac{c}{a+b}}\right)\)
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\(A=\frac{x}{x^2-2x}-\frac{x^2+4x}{x^3-4x}-\frac{2}{x^2+2x}\)(ĐK: \(x\ne0,x\ne\pm2\))
\(A=\frac{x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+4x}{x^3-4x}-\frac{2\left(x-2\right)}{x\left(x+2\right)}\)
\(A=\frac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
ĐKXĐ : x ≠ ±1/3
Ta có : \(\frac{3x-1}{6x+2}-\frac{3x+1}{2-6x}-\frac{6x}{9x^2-1}\)
\(=\frac{3x-1}{6x+2}+\frac{3x+1}{6x-2}-\frac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{\left(3x-1\right)\left(3x-1\right)}{2\left(3x-1\right)\left(3x+1\right)}+\frac{\left(3x+1\right)\left(3x+1\right)}{2\left(3x-1\right)\left(3x+1\right)}-\frac{12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{9x^2-6x+1}{2\left(3x-1\right)\left(3x+1\right)}+\frac{9x^2+6x+1}{2\left(3x-1\right)\left(3x+1\right)}-\frac{12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{2\left(9x^2-6x+1\right)}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\frac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}=\frac{3x-1}{3x+1}\)
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}=\frac{x}{x-y}-\frac{1}{x-y}+\frac{y-1}{x-y}\)
\(=\frac{x-1+y-1}{x-y}=\frac{x+y-2}{x-y}\)
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)
ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\x\ne y\end{cases}}\)
\(=\frac{x}{x-y}-\frac{1}{x-y}-\frac{y-1}{x-y}\)
\(=\frac{x-1-y+1}{x-y}\)
\(=\frac{x-y}{x-y}=1\)
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)
ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\x\ne y\end{cases}}\)
\(=\frac{x}{x-y}-\frac{1}{x-y}-\frac{y-1}{x-y}\)
\(=\frac{x-1-y+1}{x-y}\)
\(=\frac{x-y}{x-y}=1\)
\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}=\frac{x-1}{x-y}+\frac{1-y}{x-y}=\frac{x-1+1-y}{x-y}=\frac{x-y}{x-y}=1\)
\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}\)
\(\frac{x+1}{2x-2}+\frac{x-1}{2x+2}+\frac{x^2}{1-x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{x^2-1}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\frac{2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2+\left(x-1\right)^2-2x^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{\left(x-1\right)\left(x+1\right)}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne y\end{cases}}\)
\(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x^2-xy}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\)
\(=\frac{x^2}{xy\left(x-y\right)}-\frac{y\left(2x-y\right)}{xy\left(x-y\right)}=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}\)
\(=\frac{x^2-\left(2xy-y^2\right)}{xy\left(x-y\right)}=\frac{x^2-2xy+y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\) (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)+\left(x+2y\right)+4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
3x^2-4x^2+2
=-x^2+2
= - ( x^2 -2 )
=\(-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)