Bài 1:

a: \(\frac{y-1}{y+1}=\frac{\left.\left(y-1\right)\left(y-1\right)\right.}{\left(y+1\right)\left(y-1\right)}=\frac{\left(y-1\right)^2}{\left(y+1\right)\left(y-1\right)}\)

\(\frac{y+1}{y-1}=\frac{\left(y+1\right)\left(y+1\right)}{\left(y-1\right)\left(y+1\right)}=\frac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)}\)

\(\frac{1}{y^2-1}=\frac{1}{\left(y-1\right)\left(y+1\right)}\)

b: \(\frac{2}{y^2-4y}=\frac{2}{y\left(y-4\right)}=\frac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)}=\frac{2y+8}{y\left(y-4\right)\left(y+4\right)}\)

\(\frac{y}{y^2-16}=\frac{y}{\left(y-4\right)\left(y+4\right)}=\frac{y\cdot y}{y\left(y-4\right)\left(y+4\right)}=\frac{y^2}{y\left(y-4\right)\left(y+4\right)}\)

c: \(\frac{x}{x^3+1}=\frac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x-1}{x+1}=\frac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\frac{x+2}{x^2-x+1}=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)

d: \(\frac{2}{x^2+5x}=\frac{2}{x\left(x+5\right)}=\frac{2\left(x+5\right)}{x\left(x+5\right)^2}=\frac{2x+10}{x\left(x+5\right)^2}\)

\(\frac{x+5}{x^2+10x+25}=\frac{x+5}{\left.\left(x+5\right)^2\right.}=\frac{x\left(x+5\right)}{x\left(x+5\right)^2}\)

\(\frac{x+2}{x}=\frac{\left(x+2\right)\left(x+5\right)^2}{x\left(x+5\right)^2}\)

e: \(\frac{2x-3xy}{2xy}=\frac{x\left(2-3y\right)}{2xy}\)

\(\frac{x+2y}{x}=\frac{2y\left(x+2y\right)}{2xy}=\frac{2xy+4y^2}{2xy}\)

\(\frac{x-5y}{y}=\frac{2x\left(x-5y\right)}{2x\cdot y}=\frac{2x^2-10xy}{2xy}\)

f: \(\frac{x}{x^3-xy^2}=\frac{x}{x\left(x^2-y^2\right)}=\frac{1}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(\frac{1}{\left(x+y\right)^2}=\frac{1\cdot\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}=\frac{\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}\)

\(\frac{1}{\left(x-y\right)^2}=\frac{1\cdot\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}=\frac{\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

g: \(\frac{x}{x^3-4x}=\frac{x}{x\left(x^2-4\right)}=\frac{1}{x^2-4}=\frac{1}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{5}{2-x}=\frac{-5}{x-2}=\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5x-10}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{3x+1}{x+2}=\frac{\left(3x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x^2-5x-2}{\left(x+2\right)\left(x-2\right)}\)

h: \(\frac{3x}{x^3-1}=\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{x+2}{x-1}=\frac{\left(x+2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{3-x}{x^2+x+1}=\frac{\left(3-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

i: \(\frac{3}{x^2+6x+9}=\frac{3}{\left(x+3\right)^2}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)^2}=\frac{3x-9}{\left(x-3\right)\left(x+3\right)^2}\)

\(\frac{x-1}{x+3}=\frac{\left(x-1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x_{}^2-9\right)}{\left(x-3\right)\left(x+3\right)^2}\)

\(\frac{2x-6}{x^2-9}=\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)^2}\)

Bài 1:

a: \(\frac{y-1}{y+1}=\frac{\left.\left(y-1\right)\left(y-1\right)\right.}{\left(y+1\right)\left(y-1\right)}=\frac{\left(y-1\right)^2}{\left(y+1\right)\left(y-1\right)}\)

\(\frac{y+1}{y-1}=\frac{\left(y+1\right)\left(y+1\right)}{\left(y-1\right)\left(y+1\right)}=\frac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)}\)

\(\frac{1}{y^2-1}=\frac{1}{\left(y-1\right)\left(y+1\right)}\)

b: \(\frac{2}{y^2-4y}=\frac{2}{y\left(y-4\right)}=\frac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)}=\frac{2y+8}{y\left(y-4\right)\left(y+4\right)}\)

\(\frac{y}{y^2-16}=\frac{y}{\left(y-4\right)\left(y+4\right)}=\frac{y\cdot y}{y\left(y-4\right)\left(y+4\right)}=\frac{y^2}{y\left(y-4\right)\left(y+4\right)}\)

c: \(\frac{x}{x^3+1}=\frac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x-1}{x+1}=\frac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\frac{x+2}{x^2-x+1}=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)

d: \(\frac{2}{x^2+5x}=\frac{2}{x\left(x+5\right)}=\frac{2\left(x+5\right)}{x\left(x+5\right)^2}=\frac{2x+10}{x\left(x+5\right)^2}\)

\(\frac{x+5}{x^2+10x+25}=\frac{x+5}{\left.\left(x+5\right)^2\right.}=\frac{x\left(x+5\right)}{x\left(x+5\right)^2}\)

\(\frac{x+2}{x}=\frac{\left(x+2\right)\left(x+5\right)^2}{x\left(x+5\right)^2}\)

e: \(\frac{2x-3xy}{2xy}=\frac{x\left(2-3y\right)}{2xy}\)

\(\frac{x+2y}{x}=\frac{2y\left(x+2y\right)}{2xy}=\frac{2xy+4y^2}{2xy}\)

\(\frac{x-5y}{y}=\frac{2x\left(x-5y\right)}{2x\cdot y}=\frac{2x^2-10xy}{2xy}\)

f: \(\frac{x}{x^3-xy^2}=\frac{x}{x\left(x^2-y^2\right)}=\frac{1}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(\frac{1}{\left(x+y\right)^2}=\frac{1\cdot\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}=\frac{\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}\)

\(\frac{1}{\left(x-y\right)^2}=\frac{1\cdot\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}=\frac{\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

g: \(\frac{x}{x^3-4x}=\frac{x}{x\left(x^2-4\right)}=\frac{1}{x^2-4}=\frac{1}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{5}{2-x}=\frac{-5}{x-2}=\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5x-10}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{3x+1}{x+2}=\frac{\left(3x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x^2-5x-2}{\left(x+2\right)\left(x-2\right)}\)

h: \(\frac{3x}{x^3-1}=\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{x+2}{x-1}=\frac{\left(x+2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\frac{3-x}{x^2+x+1}=\frac{\left(3-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

i: \(\frac{3}{x^2+6x+9}=\frac{3}{\left(x+3\right)^2}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)^2}=\frac{3x-9}{\left(x-3\right)\left(x+3\right)^2}\)

\(\frac{x-1}{x+3}=\frac{\left(x-1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x_{}^2-9\right)}{\left(x-3\right)\left(x+3\right)^2}\)

\(\frac{2x-6}{x^2-9}=\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)^2}\)

Trên đây nó ko cho đăng ảnh,mn chịu khó nhập link này vào nha:https://i.imgur.com/xQNntGH.png

dài v

a) x(x-y)+(x-y)=(x+1)(x-y)

b) 2x+2y -x(x+y)= 2(x+y)-x(x+y)=(2-x)(x+y)

đề như nào vậy bạn

nó yêu cầu tính hay phân tích

ai giúp mình với :tìm a là số tự nhiên

a *(6/7+1/4):(19/14:1/4)<a<7/3

b*  A*2 +a*1/5=8/5

mong mg giúp 

ai cứu mình với

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Vẽ hình và tìm trục đối xứng ạ!