mình ko bt làm

a)16. x² = 64

x² = 64 : 16

x² = 4

x² = 2²

⇒ x = 2

b) (5.x - 2) - 64 = -36

(5.x - 2) = -36 + 64

5.x - 2 = 28

5.x = 28 + 2

5.x = 30

x = 30 : 5

x = 6

c) (2x - 10).(5 - x) = 0

TH1: 2x - 10 = 0

         2x = 0 + 10

         2x = 10

           x = 10 : 2

           x = 5

TH2: 5 - x = 0

              x = 5 - 0

              x = 5

⇒ Vậy x = 5.

cứu tui

\(16.x^2=64\)

\(=>x^2=64:16\)

\(=>x^2=4\)

\(=>x=2\) hoặc \(x=-2\)

_______

\(\left(5x-2\right)-64=-36\)

\(=>5x-2=\left(-36\right)+64\)

\(=>5x-2=28\)

\(=>5x=28+2\)

\(=>5x=30\)

\(=>x=30:5\)

\(=>x=6\)

_______

\(\left(2x-10\right).\left(5-x\right)=0\)

TH1: \(2x-10=0\)

\(=>2x=0+10\)

\(=>2x=10\)

\(=>x=10:2\)

\(=>x=5\)

TH2: \(5-x=0\)

\(x=5-0\)

\(x=5\)

\(=>x=5\)

1,

a,1100+(-100)=1000

b,(2017)+2010=-7

c,/-102/+36=138

d,/-1002/+(-102)=900

e,(-1002)+(-102)+515=589

1/

$(x-1)^{x+10}=(x-1)^{x+8}$

$\Rightarrow (x-1)^{x+10}-(x-1)^{x+8}=0$

$\Rightarrow (x-1)^{x+8}(x^2-1)=0$

$\Rightarrow (x-1)^{x+8}=0$ hoặc $x^2-1=0$

Nếu $(x-1)^{x+8}=0\Rightarrow x-1=0\Rightarrow x=1$

Nếu $x^2-1=0\Rightarrow x^2=1=1^2=(-1)^2\Rightarrow x=1$ hoặc $x=-1$

Vậy $x=1$ hoặc $x=-1$

2/

$1^3+2^3+3^3+...+10^3=(x+1)^2$

Ta có công thức quen thuộc:

$1^3+2^3+...+n^3=(1+2+...+n)^2=\frac{[n(n+1)]^2}{4}$

Bạn có thể xem cm tại đây:

https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

Khi đó:

$1^3+2^3+...+10^3=(x+1)^2$

$\Rightarrow \frac{[10(10+1)]^2}{4}=(x+1)^2$

$\Rightarrow 3025=(x+1)^2$

$\Rightarrow x+1=55$ hoặc $x+1=-55$

$\Rightarrow x=54$ hoặc $x=-56$

\(...\Rightarrow2x=16-10\Rightarrow2x=6\Rightarrow x=3\)

10(x-20)=10

    x - 20 = 10-10

    x - 20 = 0

    x      = 0 + 20

     x     = 20

597 - 3x = 9.2

597 -3x = 18

      3x   = 597 - 18

      3x  = 579

       x   = 579 : 3

       x   = 193

10 + 2x = 1024 : 64

10 + 2x = 16

       2x  = 16 - 10

       2x  = 6

         x  = 6 : 2

         x  = 3

10.(17 - x) - 64 = 36
<=> 10.(17 - x) = 100
<=> 17 - x = 10
<=> x = 7
@Van Dong Pham

\(10\left(17-x\right)-64=36\)

\(10\left(17-x\right)=36+64\)

\(10\left(17-x\right)=100\)

\(17-x=100:10\)

\(17-x=10\)

\(x=17-10\)

\(x=7\left(tm\right)\)

Vậy ...............

\(10\left(17-x\right)-64=36\)

\(\Leftrightarrow10\left(17-x\right)=36+64\)

\(\Leftrightarrow10\left(17-x\right)=100\)

\(\Leftrightarrow17-x=100\div10\)

\(\Leftrightarrow17-x=10\)

\(\Leftrightarrow x=17-10\)

\(\Leftrightarrow x=7\)

Vậy \(x=7\)

10(17 - x) - 64 = 36
<=> 10(17 - x) = 100
<=> 17 - x = 10
<=> x = 7
@Van Dong Pham

\(10\left(17-x\right)-64=36\)

\(10\left(17-x\right)=36+64\)

\(10\left(17-x\right)=100\)

\(17-x=100:10\)

\(17-x=10\)

\(x=17-10\)

\(x=7\left(tm\right)\)

Vậy ...............

Ta có :10.(17-x)-64=36

=> 10.(17-x)=36+64

=> 10.(17-x)=100

=> 17-x=10

=> x=7

Vậy x=7