giúp mình bài 7 với ạ

Giải thích các bước giải:

1.Ta có:

$A=\dfrac{3^{16}\cdot 5^4+3^6\cdot 5^7}{3^{14}\cdot 5^3+3^4\cdot 5^8}+\dfrac{\sqrt{100}+3\sqrt{49}-1}{3+2\sqrt{36}-\sqrt{25}}$

$\to A=\dfrac{3^6\cdot 5^4(3^{10}+5^3)}{3^4\cdot 5^3\cdot (3^{10}+ 5^5)}+\dfrac{10+3\cdot7-1}{3+2\cdot 6-5}$

$\to A=\dfrac{3^2\cdot 5(3^{10}+5^3)}{ 3^{10}+ 5^5}+3$

2.Ta có:

$\dfrac{b+c+1}a=\dfrac{a+c+2}b=\dfrac{a+b-3}c=\dfrac1{a+b+c}=\dfrac{b+c+1+a+c+2+a+b-3}{a+b+c}=\dfrac{2(a+b+c)}{a+b+c}=2$

$\to \begin{cases}b+c+1=2a\\ a+c+2=2b\\ a+b-3=2c\\ a+b+c=\dfrac12\end{cases}$

$\to \begin{cases}a+b+c+1=3a\\ a+b+c+2=3b\\ a+b+c-3=3c\\ a+b+c=\dfrac12\end{cases}$

$\to \begin{cases}\dfrac12+1=3a\\\dfrac12+2=3b\\ \dfrac12-3=3c\\ a+b+c=\dfrac12\end{cases}$

$\to \begin{cases}a=\dfrac12\\b=\dfrac56\\c=-\dfrac56\end{cases}$

$\to M=(2\cdot\dfrac12+8\cdot \dfrac56-4\cdot (-\dfrac56)-13)^3=(-2)^3=-8$