dấu [] là giá trị tuyệt đối nha

1/ 3-2x+4+6x=x+7+3x

⇔-2x+6x-x-3x=0

⇔0x=0 (Vô số nghiệm)

2/-6(1,5-2x)=3(-15+2x)

⇔-9+12x=-45+6x

⇔6x+36=0

⇔6(x+6)=0

⇔x+6=0

⇔x=-6

Vậy S ϵ {-6}

3/ 3(2x-5)+5(x-1)=4(x+1)

⇔6x-15+5x-5=4x+4

⇔7x=24

⇔x=\(\dfrac{24}{7}\) 

Vậy S ϵ {\(\dfrac{24}{7}\)}

1) Ta có: \(3-2x+4+6x=x+7+3x\)

\(\Leftrightarrow4x+7=4x+7\)

\(\Leftrightarrow4x+7-4x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

2) Ta có: \(-6\cdot\left(1.5-2x\right)=3\left(-15+2x\right)\)

\(\Leftrightarrow-9+12x=-45+6x\)

\(\Leftrightarrow12x-9+45-6x=0\)

\(\Leftrightarrow6x+36=0\)

\(\Leftrightarrow6x=-36\)

hay x=-6

Vậy: S={-6}

3) Ta có: \(3\left(2x-5\right)+5\left(x-1\right)=4\left(x+1\right)\)

\(\Leftrightarrow6x-15+5x-5=4x+4\)

\(\Leftrightarrow11x-20-4x-4=0\)

\(\Leftrightarrow7x-24=0\)

\(\Leftrightarrow7x=24\)

\(\Leftrightarrow x=\dfrac{24}{7}\)

Vậy: \(S=\left\{\dfrac{24}{7}\right\}\)

1) 3 - 2x + 4 + 6x = x + 7 + 3x

⇔-2x + 6x - x - 3x = 7 - 3 - 4

⇔0x = 0

⇔x ∈ R

Vậy ...

2) -6(1,5 - 2x) = 3(-15 + 2x)

⇔-9 + 12x = -45 + 6x

⇔6x = -36

⇔x = -6

Vậy ...

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

a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

1/ ( x-1) (2x+1) =0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)

2/ x (2x-1) (3x+15) =0

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)

3/ (2x-6) (3x+4).x=0

\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4/ (2x-10)(x²+1)=0

\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)

5/ (x²+3) (2x-1) =0

\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)

6/ (3x-1) (2x² +1)=0

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)

1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)

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

2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)

3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4: Ta có: \(\left(2x-10\right)\left(x^2+1\right)=0\)

mà \(x^2+1>0\forall x\)

nên 2x-10=0

hay x=5

5: Ta có: \(\left(x^2+3\right)\left(2x-1\right)=0\)

mà \(x^2+3>0\forall x\)

nên 2x-1=0

hay \(x=\dfrac{1}{2}\)

6: Ta có: \(\left(3x-1\right)\left(2x^2+1\right)=0\)

mà \(2x^2+1>0\forall x\)

nên 3x-1=0

hay \(x=\dfrac{1}{3}\)

1)  (2x + 1)(3x – 2) = (5x – 8)(2x + 1)

⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0

⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0

⇔ (2x + 1).(3x – 2 – 5x + 8) = 0

⇔ (2x + 1)(6 – 2x) = 0

⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)

Vậy.....

2)  4x² -1 = (2x + 1)(3x - 5)

⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0

⇔ (2x+1)(2x-1-3x+5)=0

⇔ (2x+1)(4-x)=0

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

Vậy...

3)  

(x + 1)² = 4(x² – 2x + 1)

⇔ (x + 1)² - 4(x² – 2x + 1) = 0

⇔ x² + 2x +1- 4x² + 8x – 4 = 0

⇔ - 3x² + 10x – 3 = 0

⇔ (- 3x² + 9x) + (x – 3) = 0

⇔ -3x (x – 3)+ ( x- 3) = 0

⇔ ( x- 3) ( - 3x + 1) = 0

⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy......

4) 2x³+5x²-3x=0

⇒2x³-x²+6x²-3x=0

⇒(2x³-x²)+(6x²-3x)=0

⇒x²(2x-1)+3x(2x-1)=0

⇒(x²+3x)(2x-1)=0

⇒ hoặc x²+3x=0⇒x(x+3)=0⇒hoặc x=0 hoặc x=-3

hoặc 2x-1=0⇒x=0,5

Vậy ...

5)2x=3x-2

⇒2x-3x=-2

⇒-x=-2

⇒x=2

6) x+15=3x-1

⇒x-3x=-1-15

⇒-2x=-16

⇒x=8

7)2-x=0,5x-4

⇒-x-0,5x=-4-2

⇒-1,5x=-6

⇒x=4

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15

<=> 40x = 40 <=> x = 1

Cái này mới chuẩn

b) (2x+1)(1-2x)+(1-2x)²=18 <=> 1 - 4x² + 4x² - 4x + 1 = 18

<=> -4x = 16 <=> x = -4

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

x^3=125

x=5

x⁶:x³=125

x^6-3=125

x³=125

x³=5³

x=5

vậy x=5

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6