\(\frac{a}{b}=\frac{10}{3}\)

\(\Rightarrow a=\frac{10}{3}b\)

\(M=\frac{16a^2-40ab}{8a^2-24ab}=\frac{8a\left(2a-5b\right)}{8a\left(a-3b\right)}=\frac{\frac{20}{3}b-5b}{\frac{10}{3}b-3b}=\frac{\frac{5}{3}b}{\frac{1}{3}b}=5\)

5) gọi phân số trên là D
 
Ta có : 
 

Ta có

\(M=\frac{16a^2-40ab}{8a^2-24ab}=\frac{16.\frac{a^2}{b^2}-40.\frac{a}{b}}{8.\frac{a^2}{b^2}-24.\frac{a}{b}}\)

\(=\frac{16.\left(\frac{10}{3}\right)^2-40.\frac{10}{3}}{8.\left(\frac{10}{3}\right)^2-24.\frac{10}{3}}=5\)

?????

Ta có :

\(3a=10b\Rightarrow\dfrac{a}{10}=\dfrac{b}{3}\)

Đặt \(\dfrac{a}{10}=\dfrac{b}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)

\(\Rightarrow\dfrac{16a^2-40ab}{8a^2-24ab}=\dfrac{8a\left(2a-5b\right)}{8a\left(a-3b\right)}=\dfrac{2a-5b}{a-3b}=\dfrac{2.10k-5.3k}{10k-3.3k}=\dfrac{20k-15k}{10k-9k}=\dfrac{5k}{k}=5\)

Có 3a = 10b

⇒ \(a=\frac{10b}{3}\)

Khi đó

\(\frac{16a^2-4ab}{8a^2-24ab}=\frac{4a\left(4a-b\right)}{8a\left(a-3b\right)}=\frac{\frac{40b}{3}-\frac{3b}{3}}{2\left(\frac{10b}{3}-\frac{b}{3}\right)}=\frac{\frac{27b}{3}}{2.\frac{9b}{3}}=\frac{9b}{6b}=\frac{3}{2}\)

Ko chawcs lesm :)))