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Đk: x, y \(\ne\)0
Ta có: P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\left(\frac{x^3+\left(y^2-x^2\right)\left(x+y\right)-y^3}{xy\left(x+y\right)}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(x+y\right)^2}{xy\left(x+y\right)}\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\frac{\left(x-y\right)\left(x^2+xy+y^2-x^2-2xy-y^2\right)}{xy\left(x^2+xy+y^2\right)}\)
P = \(\frac{2}{x}-\frac{-xy\left(x-y\right)}{xy\left(x^2+xy+y^2\right)}=\frac{2}{x}+\frac{x-y}{x^2+xy+y^2}=\frac{2x^2+2xy+2y^2+x^2-xy}{x\left(x^2+xy+y^2\right)}\)
P = \(\frac{3x^2+xy+2y^2}{x\left(x^2+xy+y^2\right)}\)
b) Ta có: x2 + y2 + 10 = 2x - 6y
<=> x2 - 2x + 1 + y2 + 6y + 9 = 0
<=> (x - 1)2 + (y + 3)2 = 0
<=> \(\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
Do đó: P = \(\frac{3.1^2-3.1+2.\left(-3\right)^2}{1\left(1^2-3+\left(-3\right)^2\right)}=\frac{18}{7}\)
\(x^2+2x+y^2-6y-10=0\)
\(x^2+2x+1+y^2-6x+9=10\)
\(\left(x+1\right)^2+\left(y-3\right)^2=0\)
\(\left(x+1\right)^2=\left(y-3\right)^2=0\)
\(x+1=y-3=0\)
Vậy \(x=-1;y=3\)
\(x^2\)\(+2x+y^2\)\(-6y-10=0\)
\(x^2\)\(+2x+1+y^2\)\(-6x+9=10\)
\(\left(x+1\right)^2\)+\(\left(y-3\right)^2\)\(=0\)
\(\left(x+1\right)^2\)\(=\left(y-3\right)^2\)\(=0\)
\(x+1=y-3=0\)
Vậy: \(x=-1;y=3\)
a) \(\left(2x+1\right)^2+2\left(2x+1\right)+1\)
\(=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot1+1^2\)
\(=\left[\left(2x+1\right)+1\right]^2\)
\(=\left(2x+2\right)^2\)
b) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y\right)^2\)
\(=4y^2\)
\(x^2+y^2+2x-6y+10=\left(x^2+2x+1\right)+\left(y^2-6y+9\right)\)
\(\left(x+1\right)^2+\left(y-3\right)^2\)
neu co sai bn thong cam
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
cái này là hằng đẳng thức đấy bn ạ