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

b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(7-5\right)=-2\)

c) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

d) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}+\left(-\sqrt{5}\right)\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-2\)

128.912 = 1816

Ta có: 128.912 = (4.3)8.912 =48.38.912 =(22)8.(32)4.912

= 216.94.912 = 216.916= (2.9)16 = 1816

Vế trái bằng vế phải nên đẳng thức được chứng minh

a 3 b 3 = a 3 3 . b 3 3 = a b 3

Vế trái bằng vế phải nên đẳng thức được chứng minh.

\(-\sqrt{121a^2}+3\sqrt{36a^2}+\sqrt{49a^2}=-11\left|a\right|+18\left|a\right|+7\left|a\right|=11a-18a-7a=-14a\left(đpcm\right)\left(do.a< 0\right)\)

\(\dfrac{3-x}{3+x}=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{x^2-6x+9}{9-x^2}\)