Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+2}{4x-x^2-3}=\frac{-\left(x+2\right)}{x^2-4x+3}=\frac{\left(-x-2\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}=\frac{-2x^2-9x-10}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)
\(\frac{1}{2x^2+3x-5}=\frac{1}{\left(x-1\right)\left(2x+5\right)}=\frac{x-3}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)
\(\dfrac{4}{x^2-9}=\dfrac{4x}{x\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1-x}{3x-x^2}=\dfrac{x-1}{x\left(x-3\right)}=\dfrac{\left(x-1\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)
Ta có \(\frac{2}{x^3-y^3}=\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x-1}{x^2-y^2}=\frac{2x+1}{\left(x+y\right)\left(x-y\right)}\)
\(\frac{1}{x+y}\) giữ nguyên
MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)
Ta có:
\(\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{2.\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{1}{x+y}=\frac{1.\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{2x+1}{\left(x+y\right)\left(x-y\right)}=\frac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\frac{-3}{x^2+6x+8}=\frac{-3}{x\left(x+2\right)+4\left(x+2\right)}=\frac{-3}{\left(x+2\right)\left(x+4\right)}=\frac{-3x+12}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)
\(\frac{5}{x^2-16}=\frac{5}{\left(x-4\right)\left(x+4\right)}=\frac{5x+10}{\left(x+2\right)\left(x-4\right)\left(x+4\right)}\)
\(\frac{1}{x^2-2x-8}=\frac{1}{x\left(x-4\right)+2\left(x-4\right)}=\frac{1}{\left(x-4\right)\left(x+2\right)}=\frac{x+4}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)
a) \(\dfrac{3x}{2x+4}\) và \(\dfrac{x+3}{x^2-4}\)
Phân tích các mẫu thức thành nhân tử :
\(2x+4 = 2(x+2)\)
\(x^2 - 4 = (x-2)(x+2)\)
MTC : \(2(x+2)(x-2)\)
Nhân tử phụ của mẫu thức : \(2x + 4\) là \((x - 2)\)
\(x^2 - 4\) là \(2\)
QĐ: \(\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{x+3}{x^2-4}=\dfrac{x+3}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}\)
b) \(\dfrac{x+5}{x^2+4x+4}\) và \(\dfrac{x}{3x+6}\)
Phân tích các mẫu thức thành nhân tử :
\(x^2+4x+4 = (x+2)^2\)
\(3x + 6\) \(= 3(x+2)\)
MTC : \(3(x+2)^2\)
Nhân tử phụ của mẫu thức : \(x^2 + 4x +4 \) là \(3\)
\(3x + 6\) là \((x+2)\)
QĐ : \(\dfrac{x+5}{x^2+4x+4}=\dfrac{\left(x+5\right)}{\left(x+2\right)^2}=\dfrac{3\left(x+5\right)}{3\left(x+2\right)^2}\)
\(\dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}\)
Bài giải
a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}\) giữ nguyên.
c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
\(\frac{4}{x^2-9}=\frac{4}{\left(x-3\right)\left(x+3\right)}=\frac{4x}{x\left(x-3\right)\left(x+3\right)}\)
\(\frac{1-x}{3x-x^2}=\frac{x-1}{x^2-3x}=\frac{\left(x-1\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)