Bài 13

1) 2ˣ = 4³.4⁴

2ˣ = 4⁷

2ˣ = (2²)⁷

2ˣ = 2¹⁴

x = 14

2) 2ˣ = 4⁶.16³

2ˣ = (2²)⁶.(2⁴)³

2ˣ = 2¹².2¹²

2ˣ = 2²⁴

x = 24

3) 2ˣ = 4⁵.16²

2ˣ = (2²)⁵.(2⁴)²

2ˣ = 2¹⁰.2⁸

2ˣ = 2¹⁸

x = 18

4) 2ˣ = 2⁵.2⁶

2ˣ = 2¹¹

x = 11

5) 2ˣ = 4⁴.2³

2ˣ = (2²)⁴.2³

2ˣ = 2⁸.2³

2ˣ = 2¹¹

x = 11

6) 2ˣ = 16⁵.32³

2ˣ = (2⁴)⁵.(2⁵)³

2ˣ = 2²⁰.2¹⁵

2ˣ = 2³⁵

x = 35

Bài 14:

1) 5.3ˣ = 5.3⁴

3ˣ = 5.3⁴:5

3ˣ = 3⁴

x = 4

2) 7.4ˣ = 7.4³

4ˣ = 7.4³:7

4ˣ = 4³

x = 3

3) 3/5 . 4ˣ = 3/5 . 4⁵

4ˣ = 3/5 . 4⁵ : 3/5

4ˣ = 4⁵

x = 5

4) 3/2 . 5ˣ = 3/2 . 5¹²

5ˣ = 3/2 . 5¹² : 3/2

5ˣ = 5¹²

x = 12

5) 8.7ˣ = 8.7⁶

7ˣ = 8.7⁶ : 8

7ˣ = 7⁶

x = 6

6) 2ˣ = 2.2⁸

2ˣ = 2⁹

x = 9

7) 5ˣ = 5⁴.5⁸

5ˣ = 5¹²

x = 12

8) 5.3ˣ = 7.3⁵ - 2.3⁵

5.3ˣ = 3⁵.(7 - 2)

5.3ˣ = 3⁵.5

3ˣ = 3⁵.5 : 5

3ˣ = 3⁵

x = 5

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