THPT : 15xy^2(x-y)/25xy(y-x)
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a, =-3y/5
b,A=x2-2x.1/2+1/4+3/4=(x-1/2)2+3/4 > hoặc=3/4 suy ra >0 với mọi x thuộc R
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a,\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\dfrac{2y}{3\left(x+y\right)^2}\)
b,\(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-x\right)+\left(-xy+y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\dfrac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\dfrac{x-y}{x+y}\)
c,\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left[\left(x-2\right)\left(x^2+2x+4\right)\right]}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
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\(12x^2+13y^2=25xy\)
\(\Leftrightarrow12x^2-25xy+13y^2=0\)
\(\Leftrightarrow12x^2-12xy-13xy+13y^2=0\)
\(\Leftrightarrow12x\left(x-y\right)-13y\left(x-y\right)=0\)
\(\Leftrightarrow\left(12x-13y\right)\left(x-y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}12x-13y=0\\x-y=0\end{cases}}\)
Mà để A xác định \(\Leftrightarrow x-y\ne0\) Do đó \(12x-13y=0\Leftrightarrow12x=13y\Rightarrow x=\frac{13}{12}y\)
\(\Rightarrow A=\frac{\frac{13}{12}y+y}{\frac{13}{12}y-y}=\frac{y\left(\frac{13}{12}+1\right)}{y\left(\frac{13}{12}-1\right)}=\left(\frac{13}{12}+1\right):\left(\frac{13}{12}-1\right)=\frac{25}{12}:\frac{1}{12}=25\)
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\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)