Bài 8:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

Vì \(8^{75}< 9^{75}\Rightarrow2^{225}< 3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\Rightarrow2^{91}>5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

7520 = 4510.530

Ta có: 4510.530 = (9.5)10.530 = 910.510.530 = (32)10.540

=320.(52)20 = 320.2520 = (3.25)20 = 7520

Vế phải bằng vế trái nên đẳng thức được chứng minh

a: \(45^{10}\cdot5^{30}=45^{10}\cdot\left(5^3\right)^{10}\)

\(=\left(45\cdot5^3\right)^{10}=\left(3^2\cdot5^4\right)^{10}=3^{20}\cdot5^{40}\)

\(75^{19}=\left(5^2\cdot3\right)^{19}=5^{38}\cdot3^{19}\)

\(\frac{45^{10}\cdot5^{30}}{75^{19}}=\frac{3^{20}\cdot5^{40}}{5^{38}\cdot3^{19}}=5^2\cdot3=75>1\)

Do đó: \(45^{10}\cdot5^{30}>75^{19}\)

=>Đúng

b: Đúng

c: \(99^{20}=\left(99^2\right)^{10}=9801^{10}<9999^{10}\)

=>Sai

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

=(x+1)/(2x-1)

Đề ạ

\(sin^6a+cos^6a=\left(sin^2a\right)^3+\left(cos^2a\right)^3\)

\(=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

\(=1-3sin^2a.cos^2a\)