a: \(\frac{x+y}{x^2\left(y+z\right)}=\frac{\left(x+y\right)\cdot y^2z^2\left(x+z\right)\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{\left(x+z\right)\cdot y^2z^2\left(x+y\right)^2}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

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

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

b: \(\frac{5x}{x^2+5x+6}=\frac{5x}{\left(x+2\right)\left(x+3\right)}=\frac{5x\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\frac{2x+3}{x^2+7x+10}=\frac{2x+3}{\left(x+2\right)\left(x+5\right)}=\frac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

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

Giúp mk vs các bạn ới!

a: \(\dfrac{x-1}{x+1}=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\)

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

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

b: \(\dfrac{x}{x^3-xy^2}=\dfrac{1}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)^2}\)

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

c: \(\dfrac{5x^2}{x^2+5x+6}=\dfrac{5x^2}{\left(x+2\right)\left(x+3\right)}=\dfrac{5x^2\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\dfrac{2x+3}{x^2+7x+10}=\dfrac{2x+3}{\left(x+2\right)\left(x+5\right)}=\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(-5=\dfrac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

Ta có mẫu thức chung phải chia hết cho từng mẫu thức riêng.

Vì phép chia này là phép chia hết nên số dư phải bằng 0, tức là:

3 – a(4 – a) = 0 và 2 – 2a = 0 ⇒ a = 1.

Vậy phân thức thứ nhất là 

Vì phép chia này là phép chia hết nên số dư phải bằng 0, tức là:

6 – b = 0 và -6 + b = 0 ⇒ b = 6.

Vậy phân thức thứ hai là 

* Quy đồng:

Tại sao lại là x+2 và x-2

\(\dfrac{5x^2}{x^2+5x+6}=\dfrac{5x^2}{\left(x+2\right)\left(x+3\right)}=\dfrac{5x^2\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\dfrac{2x+3}{x^2+7x+10}=\dfrac{2x+3}{\left(x+2\right)\left(x+5\right)}=\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(-5=\dfrac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

qui đồng cái nào hả bn ?? ?

MSC:

\(\dfrac{5x^2}{\left(x+2\right)\left(x+3\right)};\dfrac{\left(2x+3\right)}{\left(x+2\right)\left(x+5\right)};-5\)

\(\dfrac{5x^2\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)};\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)};\dfrac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)