Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{3y}{4x}+\frac{5y}{4x}=\frac{3y+5y}{4x}=\frac{8y}{4x}=\frac{2y}{x}\)
b)\(\frac{x^2+1}{2x-4}-\frac{7x}{2-x}=\frac{x^2+1}{2\left(x-2\right)}-\frac{-7x}{x-2}\)
\(=\frac{x^2+1}{2\left(x-2\right)}-\frac{-7x\times2}{\left(x-2\right)\times2}=\frac{x^2+1+14x}{2\left(x-2\right)}\)
a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)
b, Sua de : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{1}{x}\)
Câu 1: xin sửa đề :D
CM: \(n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)là 1 scp
\(n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)
\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)
\(=\left(n^2+3n\right)^2+2\left(n^2+3n\right)+1\)
\(=\left(n^2+3n+1\right)^2\)là scp
\(\frac{x^2+2}{2xy^3}-\frac{2x+2}{2xy^3}=\frac{x^2+2-2x-2}{2xy^3}=\frac{x^2-2x}{2xy^3}=\frac{x\left(x-2\right)}{2xy^3}=\frac{x-2}{2y^3}\)
\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{25-x^2}=\frac{4}{x-5}-\frac{1}{x+5}+\frac{x^2-13x}{x^2-25}\)
\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-13x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{4x+20-x+5+x^2-13x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)