Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
a.
\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\) (đề thế này đúng ko? :D)
\(=\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}\)
\(=\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+6x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}=\dfrac{x+3}{x-3}\)
b.
\(\left(x+2\right)^2-\left(x+1\right)\left(x-1\right)=\left(x^2+4x+4\right)-\left(x^2-1\right)=4x+6\)
c.
\(\dfrac{5x^2-10xy}{x^2-4xy+4y^2}\)
\(=\dfrac{5x\left(x-2y\right)}{x^2-2\cdot x\cdot2y+\left(2y\right)^2}=\dfrac{5x\left(x-2y\right)}{\left(x-2y\right)^2}=\dfrac{5x}{x-2y}\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
a: \(=\dfrac{5\left(x^2+2xy+y^2\right)}{3\left(x^3+y^3\right)}\)
\(=\dfrac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)
b: \(=\dfrac{x^2-4xy+4y^2-4}{2x\left(x-2y+2\right)}=\dfrac{\left(x-2y-2\right)\left(x-2y+2\right)}{2x\left(x-2y+2\right)}\)
\(=\dfrac{x-2y-2}{2x}\)
c: \(=\dfrac{2\left(x^2+5x+1\right)}{x\left(x-2\right)\left(x+2\right)}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)