\(P=\dfrac{x^3-y^3}{x^2y-xy^2}-\dfrac{x^3+y^3}{x^2y+xy^2}-\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{x+y}{x-y}-\dfrac{x-y}{x+y}\right)\)

\(=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x-y\right)}-\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy\left(x+y\right)}-\dfrac{x^2-y^2}{xy}\cdot\dfrac{x^2+2xy+y^2-x^2+2xy-y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{x^2+xy+y^2-x^2+xy-y^2}{xy}-\dfrac{\left(x-y\right)\left(x+y\right)}{xy}\cdot\dfrac{4xy}{\left(x-y\right)\left(x+y\right)}\)

\(=2-4=-2\)

Bài 2: 

a: Ta có: \(A=\left(x^2-3x+5\right)-\left(x^2+4x-1\right)+5x^2-3\)

\(=x^2-3x+5-x^2-4x+1+5x^2-3\)

\(=5x^2-7x+3\)

b: Ta có: \(B=\left(3x^2-11x+7\right)-\left(2x^2+3x+4\right)\)

\(=3x^2-11x+7-2x^2-3x-4\)

\(=x^2-14x+3\)

\(\Rightarrow P=\left(\dfrac{\left(y-x\right)\left(y+x\right)}{y-x}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x+y\right)}\right).\dfrac{x+y}{y^2-2xy+x^2+xy}\)

\(\Rightarrow P=\left(y+x-\dfrac{x^2+xy+y^2}{x+y}\right).\dfrac{x+y}{y^2-xy+x^2}\)

\(\Rightarrow P=\dfrac{\left(x+y\right)^2-\left(x^2+xy+y^2\right)}{x+y}.\dfrac{x+y}{y^2-xy+x^2}\)

\(\Rightarrow P=\dfrac{x^2+2xy+y^2-x^2-xy-y^2}{x+y}.\dfrac{x+y}{y^2-xy+x^2}\)

\(\Rightarrow P=\dfrac{xy}{x+y}.\dfrac{x+y}{y^2-xy+x^2}\)

\(\Rightarrow P=\dfrac{xy}{y^2-xy+x^2}\)

\(\left(x+3\right)\left(x^2-3x+3^2\right)-x\left(x^2-3\right)\)

\(=x^3+3^3-x^3-3x\)

\(=3^3-3x\)

\(=3\left(3^2-x\right)\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-3\right)\)

\(=x^3+3^3-x^3+3x\)

\(=27+3x\)

\(=3\left(9+x\right)\)

\(10,\\ a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\\ \Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge0\\ \Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow\dfrac{a}{2}=b=c=d=e\)

\(4,\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{4}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\ge3ab+3bc+3ca\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{2}a^2-\dfrac{1}{2}b^2-\dfrac{1}{2}c^2-ab-bc-ac\ge0\\ \Leftrightarrow\dfrac{1}{2}a^2+\dfrac{1}{2}b^2+\dfrac{1}{2}c^2+ab+ac+bc\ge0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge0\\ \Leftrightarrow\left(a+b+c\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a+b+c=0\)

3:

a: =>(x-2)(x+2)-3(x-2)^2=0

=>(x-2)(x+2-3x+6)=0

=>(x-2)(-2x+8)=0

=>x=2 hoặc x=4

b: =>15x+10=12x+1

=>3x=-9

=>x=-3

c; =>x-3=0 hoặc 2x+4=0

=>x=-2 hoặc x=3

d: =>2/3x-1-1/4x+1/4=3

=>5/12x-3/4=3

=>5/12x=15/4

=>x=9

1h30' mik gửi đáp án

haiz( bất lực )

\(2x^2+y^2+4x-2y+3=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow2\left(x+1\right)^2+\left(y-1\right)^2=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

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