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

\(\frac{8a^2}{4a^2x-x^3}=\frac{8a^2}{x\left(4a^2-x^2\right)}=\frac{8a^2}{x\left(2a-x\right)\left(2a+x\right)}\)

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

Mn giúp mik vs ạ

a) \(MTC=a^2x^2b^2\)

\(NTP:a^2x^2b^2:a^2x=xb^2\)

\(a^2x^2b^2:x^2b=a^2b\)

\(a^2x^2b^2:b^2a=ax^2\)

Quy đồng :

\(\dfrac{a+x}{a^2x}=\dfrac{\left(a+x\right)\cdot xb^2}{a^2x.xb^2}=\dfrac{axb^2+x^2b^2}{a^2x^2b^2}\)

\(\dfrac{a+b}{x^2b}=\dfrac{\left(a+b\right)\cdot a^2b}{x^2b\cdot a^2b}=\dfrac{a^3b+a^2b^2}{a^2x^2b^2}\)

\(\dfrac{b+a}{b^2a}=\dfrac{\left(b+a\right)\cdot ax^2}{b^2a\cdot ax^2}=\dfrac{abx^2+a^2x^2}{a^2x^2b^2}\)

a: \(\dfrac{2-a}{2a}=\dfrac{2-a}{2a}\)

\(\dfrac{1}{a}=\dfrac{a}{2a}\)

b: \(\dfrac{5}{3}=\dfrac{5a}{3a}\)

\(\dfrac{5a-6}{3a}=\dfrac{5a-6}{3a}\)

Các phân thức đã được quy đồng mẫu từ các phân thức \(\frac{150}{\text{t}};\frac{2a}{\frac{1}{2}mn}\text{ và }\frac{\left(p-q\right)\cdot2}{xyz}\) là :

\(\frac{75\cdot mn\cdot xyz}{\text{t}\cdot\frac{1}{2}mn\cdot xyz};\frac{2a\cdot\text{t}\cdot xyz}{\text{t}\cdot\frac{1}{2}mn\cdot xyz}\text{ và }\frac{\left(p-q\right)\cdot mn\cdot\text{t}}{\text{t}\cdot\frac{1}{2}mn\cdot xyz}\).

Sửa lại : \(\frac{150}{\text{t}};\frac{2a}{\frac{1}{2}mn}\text{ và }\frac{\left(p-q\right)\cdot2}{xyz}\).