a) 7⁶ + 7⁵ - 7⁴

= 7⁴.(7² + 7 - 1)

= 7⁴.55 ⋮ 55

Vậy (7⁶ + 7⁵ - 7⁴) ⋮ 55

b) 81⁷ - 27⁹ + 3²⁹

= (3⁴)⁷ - (3³)⁹ + 3²⁹

= 3²⁸ - 3²⁷ + 3²⁹

= 3²⁶.(3² - 3 + 3³)

= 3²⁶.(9 - 3 + 27)

= 3²⁶.33 ⋮ 33

Vậy (81⁷ - 27⁹ + 3²⁹) ⋮ 33

(-8/9 + 7/5 - 2/11) - (-2/5 - 17/9 - 13/11)

= -8/9 + 7/5 - 2/11 + 2/5 + 17/9 + 13/11

= (-8/9 + 17/9) + (7/5 + 2/5) + (-2/11 + 13/11)

= 1 + 9/5 + 1

= 19/5

(-8/9 +7/5 - 2/11) - (-2/5-17/9-13/11)

=-8/9 +7/5 - 2/11 +2/5 +17/9 +13/11

=(-8/9 + 17/9) +(7/5+2/5) (-2/11+13/11)

=11/9 + 9/5 + 1

= 55/45 + 81/45 +45/45

=181/45

Sao mình không thấy biểu thức đâu bạn nhỉ?

Số vô tỉ là không phải số hữu tỉ không thể biểu diễn dưới dạng tỉ số của 2 số nguyên 

\(\left|5x\right|-3x=2\)

\(\text{⇒}\left|5x\right|=3x+2\)  

TH1: 

\(5x=3x+2\) \(\left(x\ge0\right)\)

\(\text{⇒}5x-3x=2\)

\(\text{⇒}2x=2\)

\(\text{⇒}x=\dfrac{2}{2}\)

\(\text{⇒}x=1\left(tm\right)\)

TH2: 

\(-5x=3x+2\) (x < 0) 

\(\text{⇒}-5x-3x=2\)

\(\text{⇒}-8x=2\)

\(\text{⇒}x=-\dfrac{1}{4}\left(tm\right)\)

Xét \(x>0\)

\(5x-3x=2\)

\(\left(5-3\right)x=2\)

\(2x=2\)

\(x=1\)

Xét \(x< 0\)

\(\left(-5x\right)-3x=2\)

\(\left(-5-3\right)x=2\)

\(-8x=2\)

\(x=-\dfrac{2}{8}=-\dfrac{1}{4}\)

Vậy: \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Ta viết lại tổng này thành:

\(P=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\right)-\dfrac{49}{99}\)

\(P=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}-\dfrac{49}{99}\right)\)

\(P=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)

\(P=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)

\(P=\dfrac{1}{2}-\dfrac{1}{198}+\dfrac{1}{4}-\dfrac{1}{200}-\dfrac{49}{99}\)

\(P=\dfrac{49}{200}\)

Ta có BĐT: \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\)

Áp dụng ta có:

\(A=\left|x-2\right|-\left|x+4\right|\le\left|x-2-x-4\right|=\left|-6\right|=6\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-2\ge0\\x+4\ge0\end{matrix}\right.\Leftrightarrow x\ge2\)

Vậy: \(A_{max}=6\Leftrightarrow x\ge2\)

$3$

Theo BĐT: \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\) ta có:

\(B=\left|2x-7\right|-\left|2x-11\right|\le\left|2x-7-2x+11\right|=\left|4\right|=4\) 

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2x-7\ge0\\2x-11\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{11}{2}\)

Vậy: \(B_{max}=4\Leftrightarrow x\ge\dfrac{11}{2}\)

a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)

\(=3^{11}\cdot2^{30}\)

\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)

Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)

Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)

b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)

\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)

Vậy dãy trên nhỏ hơn 1

a/

\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)

\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)

\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)

b/

\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)

\(=1-\dfrac{1}{10^2}< 1\)