Bài 10:
a: \(\left(-16\right)^{11}=-16^{11}=-\left(2^4\right)^{11}=-2^{44}\)
\(\left(-32\right)^9=-32^9=-\left(2^5\right)^9=-2^{45}\)
mà \(-2^{44}>-2^{45}\left(2^{44}<2^{45}\right)\)
nên \(\left(-16\right)^{11}>\left(-32\right)^9\)
b: \(10A=\frac{10^{2025}+10}{10^{2025}+1}=\frac{10^{2025}+1+9}{10^{2025}+1}=1+\frac{9}{10^{2025}+1}\)
\(10B=\frac{10^{2026}+10}{10^{2026}+1}=\frac{10^{2026}+1+9}{10^{2026}+1}=1+\frac{9}{10^{2026}+1}\)
Ta có: \(10^{2025}+1<10^{2026}+1\)
=>\(\frac{9}{10^{2025}+1}>\frac{9}{10^{2026}+1}\)
=>\(\frac{9}{10^{2025}+1}+1>\frac{9}{10^{2026}+1}+1\)
=>10A>10B
=>A>B
Bài 11:
\(P=3^{n+2}-2^{n+2}+3^{n}-2^{n}\)
\(=\left(3^{n}\cdot9+3^{n}\right)-\left(2^{n}\cdot4+2^{n}\right)\)
\(=3^{n}\cdot10-2^{n}\cdot5=3^{n}\cdot10-2^{n-1}\cdot10=10\left(3^{n}-2^{n-1}\right)\) ⋮10
Bài 12:
Ta có: \(\frac{x+121}{21}+\frac{x+144}{22}+\frac{x+169}{23}=6\)
=>\(\left(\frac{x+121}{21}-1\right)+\left(\frac{x+144}{22}-2\right)+\left(\frac{x+169}{23}-3\right)=0\)
=>\(\frac{x+100}{21}+\frac{x+100}{22}+\frac{x+100}{23}=0\)
=>\(\left(x+100\right)\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
=>x+100=0
=>x=-100
