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

\(\text{⇒}\left|5x\right|=3x+2\)  

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/

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

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

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

a) (-5/9)^10 : x = (-5/9)^8

=> x = (-5/9)^10 : (-5/9)^8

=> x = (-5/9)^10-8 = (-5/9)^2

=> x = 25/81

b ) x : (-5/9)^8 = (-9/5)^8

=> x = (-9/5)^8 . (-5/9)^8

=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )

=> x = 1

C) x^3 = -8 =(-2)^3

=> x = -2

a) (-5/9)¹⁰ : x = (-5/9)⁸

x = (-5/9)¹⁰ : (-5/9)⁸

x = (-5/9)²

x = 25/81

b) x : (-5/9)⁸ = (-9/5)⁸

x = (-9/5)⁸ . (-5/9)⁸

x = [-9/5 . (-5/9)]⁸

x = 1⁸

x = 1

c) x³ = -8

x³ = (-2)³

x = -2

a) \(\left[\left(-2,7\right)^4\right]^5-\left[\left(-2,7\right)^2\right]^{20}\)

\(=\left(-2,7\right)^{20}-\left(-2,7\right)^{20}\)

\(=0\)

b) \(\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\dfrac{17}{2}\right)^7:\left(\dfrac{17}{2}\right)^6\)

\(=\left(-0,5\right)^2-\dfrac{17}{2}\)

\(=0,25-\dfrac{17}{2}\)

\(=-8,25\)

c) \(\left(8^{14}:4^{12}\right):\left(16^6:8^2\right)\)

\(=8^{14}:4^{12}:16^6\cdot8^2\)

\(=2^{48}:2^{24}:2^{24}\)

\(=0\)

Cái cuối bằng 1 nhé!