Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = \(\frac{3.1^2-3.1+2.\left(-3\right)^2}{1\left(1^2-3+\left(-3\right)^2\right)}=\frac{18}{7}\)
\(A=x^4+2x^3+7x^2+6x+9\)
\(=\left(x^2\right)^2+2.x^2.x+x^2+6\left(x^2+x\right)+9\)
\(=\left(x^2+x\right)^2+2.\left(x^2+x\right).3+3^2\)
\(=\left(x^2+x+3\right)^2\)
2, \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\Rightarrow152=\left(x+y\right).19\)
\(\Rightarrow x+y=8\)
Mà \(x-y=2\Rightarrow\hept{\begin{cases}x=\left(8+2\right):2=5\\y=x-2=3\end{cases}}\)
Vậy x = 5 và y = 3
a )
Ta có :
\(\hept{\begin{cases}x+y=7\\x.y=12\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=49\\4xy=48\end{cases}}}\)
Mà \(\left(x-y\right)^2=\left(x+y\right)^2-4xy\)
\(\Rightarrow\left(x-y\right)^2=49-48\)
\(\Rightarrow\left(x-y\right)^2=1\)
Vậy \(\left(x-y\right)^2=1\)
b )
\(\hept{\begin{cases}x-y=10\\xy=3\end{cases}\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=100\\4xy=12\end{cases}}}\)
Mà \(\left(x+y\right)^2=\left(x-y\right)^2+4xy\)
\(\Rightarrow\left(x+y\right)^2=100+12\)
\(\Rightarrow\left(x+y\right)^2=112\)
Vậy \(\left(x+y\right)^2=112\)
c )
Ta có :
\(\hept{\begin{cases}xy=0\\x+y=-5\end{cases}\Rightarrow\hept{\begin{cases}4xy=0\\\left(x+y\right)^2=25\end{cases}}}\)
Mà \(\left(x-y\right)^2=\left(x+y\right)^2-4xy\)
\(\Rightarrow x^2-2xy+y^2=25-0\)
\(\Rightarrow x^2-xy+y^2-xy=25\)
\(\Rightarrow x^2-xy+y^2-0=25\)
\(\Rightarrow x^2-xy+y^2=25\)
Mà \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\Rightarrow x^3+y^3=-5.25\)
\(\Rightarrow x^3+y^3=-125\)
Vậy \(x^3+y^3=-125\)