Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
\(b,\dfrac{2y}{2x^2-xy}+\dfrac{4x}{xy-2x^2}\)
\(=\dfrac{2y}{2x^2-xy}-\dfrac{4x}{2x^2-xy}\)
\(=\dfrac{2y-4x}{2x^2-xy}\)
\(=\dfrac{2\left(y-2x\right)}{x\left(2x-y\right)}\)
\(=\dfrac{-2\left(2x-y\right)}{x\left(2x-y\right)}\)
\(=-\dfrac{2}{x}\)
\(c,\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}-\dfrac{2x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x^2+2x+5-4x}{x-3}\)
\(=\dfrac{-3x^2-2x+9}{x-3}\)
\(a,\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
\(b,\dfrac{2y}{2x^2-xy}+\dfrac{4x}{xy-2x^2}\)
\(=\dfrac{2y}{2x^2-xy}-\dfrac{4x}{2x^2-xy}\)
\(=\dfrac{2y-4x}{2x^2-xy}\)
\(=\dfrac{2\left(y-2x\right)}{x\left(2x-y\right)}\)
\(=\dfrac{-2\left(2x-y\right)}{x\left(2x-y\right)}\)
\(=-\dfrac{2}{x}\)
\(c,\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}-\dfrac{2x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x^2+2x+5-4x}{x-3}\)
\(=\dfrac{-3x^2-2x+9}{x-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(16x^3+0,25yz^3\)
\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)
\(=0,25\left(64x^3+yz^3\right)\)
b: \(x^4-4x^3+4x^2\)
\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)
\(=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
c: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)^2\)
d: \(x^3+x^2+x+1\)
\(=x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+1\right)\)
e: \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
f: \(2x^2-18\)
\(=2\cdot x^2-2\cdot9\)
\(=2\left(x^2-9\right)=2\left(x-3\right)\left(x+3\right)\)
g: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
h: \(x^4y^4+4\)
\(=x^4y^4+4x^2y^2+4-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)
i: \(x^4+4y^4\)
\(=x^4+4x^2y^2+4y^4-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)
k: \(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
a) \(\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)