a) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^2x-6 = 1

 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

c) Tớ ko bít 

d) (x - 6)³ = (x - 6)²

Th₁:

x - 6 = 1

x = 1 + 6

x = 7

Th₂:

x - 6 = 0

x = 6

Vậy x = 7

      x = 6

--thodagbun--

a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5

b, 4²x+6=1 <=> 16x=-5 <=>x=-5/16

c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)

TH1: 10x-20 = 0 <=> x=2

TH2: 3-x=1 <=> x=2

Vậy x=2

d, (x-6)^3 = (x-6)^2

<=> (x-6)^2.[(x-6)-1]=0

<=> (x-6)^2=0 hoặc (x-6)-1=0

<=> x=6 hoặc x=7

a) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^2x-6 = 1

<=> 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

a) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^{2x - 6} = 1

<=> 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

x2x2 là sao bn

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

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

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a, 7.7^x+1=343

   7^x+1+1=343

    7^x+2=343=7³

=> x+2=3

      x=3-2=1

b,2³.2^x=64

     2^3+x=64=2⁶

  => 3+x=6

        x=6-3=3

c,(3x-15)⁷=0

  =>  (3x-15)=0

    3x-15=0

    3x=0+15

    3x=15

     x=15:3=5

d,  4^(2x-6)=1

  => 4^(2x-6)=4⁰=1

=>    2x-6=0

        2x=0+6=6

         x=6:2=3

e, (3-x)^10x:(3-x)²⁰=1

Nx:   Một số chia cho chính nó luôn bằng1

Có:   3-x=3-x

=>  10x=20

       x=20:10=2

f, (x-6)³=(x-6)²

Ta có:       Th1:      0³=0²=0

                  =>  (x-6)³=(x-6)²=0

                   =>  x-6 =0

                        x=0+6=6

             Th2:    1³=1²=1

                    =>    (x-6)³=(x-6)²=1

                      =>   x-6=1

                              x=1+6=7