Bài 1:

a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)

b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)

c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)

d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)

e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)

Bài 2:

\(40-x=2^6\cdot2^2\)

\(\Rightarrow40-x=2^8\)

\(\Rightarrow40-x=256\)

\(\Rightarrow x=40-256\)

\(\Rightarrow x=-216\)

b) \(3^2\cdot3^x=81\)

\(\Rightarrow3^{2+x}=3^4\)

\(\Rightarrow2+x=4\)

\(\Rightarrow x=4-2=2\)

c) \(2^x=512\)

\(\Rightarrow2^x=2^9\)

\(\Rightarrow x=9\)

d) \(x^5=243\)

\(\Rightarrow x^5=3^5\)

\(\Rightarrow x=3\)

Bài 3:

a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)

b) \(8^3=\left(2^3\right)^3=2^9=512\)

c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)

d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)

e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)

\(=18-\left(4+3\right)\)

\(=18-7=11\)

2:

a: =>40-x=256

=>x=40-256=-216

b: =>x+2=4

=>x=2

c: =>2^x=2^9

=>x=9

d; =>x^5=3^5

=>x=3

350<351

207>170

199<222<299

35x<351

x = 0

207>x70

x = 1

199<xxx<299

xxx=222

Ta có: \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.3^2^{ }=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.9=351\)

\(\Rightarrow3^x.\left(1+3+9\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=351:13=27\)

\(\Rightarrow x=3\)

3 k mk nha

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───▄▄██▌█ ░Xe chở 100000000 đến đây..

▄▄▄▌▐██▌█ ░░░░░░ ░░░░░░░░░ ░░░░░░░▐\.

███████▌█▄▄▄▄▄▄ ▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄ ▄▄▌ \.

▀❍▀▀▀▀▀▀▀❍❍▀▀▀▀ ▀▀▀▀▀▀▀▀▀▀▀❍❍ ▀▀.

(+) \(x+12\)\(⋮\)\(x+5\)

\(\left(x+5\right)+7⋮x+5\)

Vì\(x+5⋮x+5\)nên \(7⋮x+5\)

hay \(x+5\in U\left(7\right)=\pm1,\pm7\)sau đó tìm ra tất cả x

(+)\(x+12⋮x-5\)

\(\left(x-5\right)+15⋮x-5\)

vì \(\left(x-5\right)⋮x-5\)

nên \(15⋮x-5\)

hay \(x-5\in U\left(15\right)=\pm1,\pm3,\pm5,\pm15\)sau đó tính x ra

(+)\(3x-5⋮x-4\)

\(\left(3x-12\right)+7⋮x-4\)

Vì\(3x-12⋮x-4\)

nên \(7⋮x-4\)

hay \(x-4\in U\left(7\right)=\pm1,\pm7\)sau đó tính ra x

x : 689 = 38726

x = 38726 x 689

x = 26682214

x : 689 = 157

x = 157 x 689

x = 108173

\((x - 3).(2y + 1) = 7\)

Ý của bạn là chỉ yc tìm mỗi vế của biến x ạ?

\(\left(x-3\right)\cdot\left(2y+1\right)\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\) 

`\Rightarrow \text {TH1:} x - 3 = 1`

`\Rightarrow x = 1 + 3`

`\Rightarrow x = 4`

`\text {TH2:} x - 3 = 7`

`\Rightarrow x = 7 + 3`

`\Rightarrow x = 10`

`\text {TH3:} x - 3 = -1`

`\Rightarrow x = -1 + 3`

`\Rightarrow x = 2`

`\text {TH4:} x - 3 = -7`

`\Rightarrow x = -7 + 3`

`\Rightarrow x = -4`

Vậy, `x \in {-4; 4; 2; 10}`

ta có                 

trường hợp 1:(x-3)=7                 

x-3=7

x=7+3

x=10

x-3=7

x=7+3

x=10

trường hợp 2:(x-3)=1                 

x-3=1

x=1+3

x=4

5; 5,01

cảm ơn bạn nha !

3x²-75=0
<=> 3x²=75
<=> x²=25
<=> x=5
2x²-98=0
<=> 2x²=98
<=>   x²=49
<=>   x=7
x²-7x=0
<=> x(x-7)=0
<=> x=0 hoặc x=7
-3x²+5x=0
x(-3x+5)=0
x=0 hoặc -3x+5=0
x=0 hoặc -3x=-5
x=0 hoặc    x=5/3
x²+4x+4=0
(x+2)²=0
x+2=0
x=-2
 

1. 3x² - 75 = 0

<=> 3x² = 75

<=> x² = 25

<=> x = \(\sqrt{25}\)

<=> x = 5

2. x² - 7x = 0

<=> x(x - 7) = 0

<=> \(\left[{}\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

3. x² - 14x + 13 = 0

<=> x² - 13x - x + 13 = 0

<=> x(x - 13) - (x - 13) = 0

<=> (x - 1)(x - 13) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)

4. 2x² - 98 = 0

<=> 2x² = 98

<=> x² = 49

<=> x = \(\sqrt{49}\)

<=> x = 7

5. -3x² + 5x = 0

<=> x(-3x + 5) = 0

<=> \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)

6. x² - 2x - 80 = 0

<=> x² + 8x - 10x - 80 = 0

<=> x(x + 8) - 10(x + 8) = 0

<=> (x - 10)(x + 8) = 0

<=> \(\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

7. x² = 81

<=> x² - 9² = 0

<=> (x - 9)(x + 9) = 0

<=> \(\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

8. x² + 4x + 4 = 0

<=> x² + 2.x.2 + 2² = 0

<=> (x + 2)² = 0

<=> 0 = 0² - (x + 2)²

<=> (0 + x + 2)(0 - x + 2) = 0

<=> (x + 2)(-x + 2) = 0

<=> \(\left[{}\begin{matrix}x+2=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

9. 4x² + 12x + 5 = 0

<=> 4x² + 2x + 10x + 5 = 0

<=> 2x(2x + 1) + 5(2x + 1) = 0

<=> (2x + 5)(2x + 1) = 0

<=> \(\left[{}\begin{matrix}2x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)