10A=\(\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)

10B =\(\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

\(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{12}+1}\) =>\(1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)

=> 10A < 10B

=> A<B

Vậy A < B

A= \(\dfrac{1}{3}-\dfrac{2}{3^2}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

3A= 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)

A + 3A = 1- \(\dfrac{1}{3}+\dfrac{1}{3^2}\) - \(\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

=> 4A < 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}\) \(\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

Đặt : B = 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

3B = 3 - 1 + \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}+.....+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)

B + 3B = 3 - \(\dfrac{1}{3^{99}}\)

4B = 3 - \(\dfrac{1}{3^{99}}\) < 3 => B < \(\dfrac{3}{4}\)

=> 4A < \(\dfrac{3}{4}\) => A < \(\dfrac{3}{16}\) ĐPCM

5.x + 12 = 2

5.x = 2-12

5.x = -10

x = -10:5

x = -2

b) \(\dfrac{2}{3}x\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

\(\dfrac{2}{3}x\) = \(\dfrac{1}{10}-\dfrac{1}{2}\)

\(\dfrac{2}{3}x\) = \(\dfrac{-2}{5}\)

x = \(\dfrac{-2}{5}:\dfrac{2}{3}\)

x = \(\dfrac{-3}{5}\)

mk

có tính sáng tạo cao đó !!!!!

đẹp lắm !!!