1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)

  Vậy ...

2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)

  Vậy ...

3) PT \(\Rightarrow3x-4-2x+5=3\)

          \(\Rightarrow x=2\)

 Vậy ...

4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

  Vậy ...

3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)

\(\Leftrightarrow3x-4-2x+5=3\)

\(\Leftrightarrow x+1=3\)

hay x=2

x(5 – 2x) + 2x(x – 1) = 15

(x.5 – x.2x) + (2x.x – 2x.1) = 15

5x – 2x2 + 2x2 – 2x = 15

(2x2 – 2x2) + (5x – 2x) = 15

3x = 15

x = 5.

Vậy x = 5.

 PT: x(5-2x)+2x(x-1)=15 
<=> 5x - 2x² + 2x² - 2x = 15 
<=> 3x = 15 
<=> x = 5.

\(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow\left(2x^2+4x-5x-10\right)-2\left(x^2-x\right)-15=0\)

\(\Leftrightarrow2x^2-x-10-2x^2+2x-15=0\)

\(\Leftrightarrow x=25\)

  (2x-5)(x2)-2x(x-1)=15

 \(2x^2-10-2x^2+2x=15\)

 \(-10+2x=15\)

 \(-8x=15\)

       \(x=-\frac{15}{8}\)

x5-x²2+2x²-2x=15

x(5-2)+x²(2-2)=15

=>x3+x²0=15

=>x3=15

=>x=15:3

=>x=5

vậy x=5

hơi dài dòng 1 chút nhưng chắc ko sao

\(\Leftrightarrow5x-2x^2+2x^2-2x=15\Leftrightarrow3x=15\Leftrightarrow x=5\)

Ta có:

x( 5 - 2x ) + 2x( x - 1 ) = 15 

=> 5x-2x²+2x²-2x=15

=> 3x=15

=> x=5

bạn chỉ cần phân tích ra là đuợc :)

\(\left(x+7\right)+3x\left(2x-1\right)-2x\left(3x+15\right)=-42\)

=>\(x+7+6x^2-3x-6x^2-30x=-42\)

=>\(-32x=-42-7=-49\)

=>\(x=\dfrac{49}{32}\)

a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)

\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)

\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)

\(\Rightarrow x=\frac{23}{20}\)

b) \(x+\frac{46}{15}=1,5\)

\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)

\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)

\(\Rightarrow x=\frac{-47}{30}\)

c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)

\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)

\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)

x(5 - 2x) + 2x(x - 1) = 15
=> 5x - 2x\(^2\)+ 2x\(^2\)- 2x = 15
=> 5x - 2x = 15
=> 3x = 15
=> x = 5
K cho mình nhé! Cảm ơn

k cho mk nhé

\(x\left(5-2x\right)+2x\left(x-1\right)=15\)

\(\Rightarrow5x-2x^2+2x^2-2x=15\)

\(\Rightarrow\left(5x-2x\right)+\left(-2x^2+2x^2\right)=15\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=15:3\)

\(\Rightarrow x=5\)

Vậy x = 5

_Chúc bạn học tốt_

`@` `\text {Ans}`

`\downarrow`

`a,`

`x^2 + 2x + 1 = 9`

`=> x^2 + 2x + 1 - 9 = 0`

`=> x^2 + 2x - 8 = 0`

`=> x^2 + 4x - 2x - 8 = 0`

`=> (x^2 + 4x) - (2x + 8) = 0`

`=> x(x + 4) - 2(x + 4) = 0`

`=> (x-2)(x+4) = 0`

`=>`\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy, `x \in {2; 4}`

`b,`

`x^2 - 1 = 15`

`=> x^2 = 15 + 1`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = +-4`

Vậy, `x \in {4; -4}`

`c)`

`19 - 2x^2 = 1`

`=> 2x^2 = 19 - 1`

`=> 2x^2 = 18`

`=> x^2 = 18 \div 2`

`=> x^2 = 9`

`=> x^2 = (+-3)^2`

`=> x = +-3`

Vậy, `x \in {3; -3}.`

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt