a,

2x-2018-x=1

=> x-2018=1

=> x=2019

b, (x+1)+(x+2)+(x+3)+...+(x+10)=165

Dãy trên có số số hạng là :

(10-1):1+1=10 ( số )

=> (x+1)+(x+2)+(x+3)+...+(x+10)=165

=> x+1+x+2+...+x+10=165

=> 10x.(1+...+10)=165

=> 10x.((10+1).10:2)=165

=> 10x.55=165

=> x=3

c, x+3.3-(20+49.2:7-18)=0

tính phần (20+49.2:7-18) ra là đc mà

a) (x - 1)² = 1.

<=> x - 1 = 1 hoặc x - 1 = -1.

<=> x = 2 hoặc x = 0.      

b) 7^{2x - 6} = 49.

<=> 7^{2x - 6} = 7².   

<=> 2x - 6 = 2.

<=> x = 4.  

c) (2x - 16)⁷ = 128.

<=> (2x - 16)⁷ = 2⁷.

<=> 2x - 16 = 2.

<=> x = 9.

a) X + 20% x X = 7,2

<=> X x ( 1+ 20%) = 7,2

<=> X x 6/5 = 7,2

<=> X = 6

b) X x 4 + 1/2 x X = 55,35

<=> X x ( 4+ 1/2) = 55,35

<=> X x 4,5 = 55,35

<=> X = 12,3

c) 40% x X + X x 20% = 72

<=> X x ( 40% + 20%) = 72

<=> X x 60% = 72

<=> X x 3/5 = 72

<=> X = 120

d) X - 20% x X = 216

<=> X x ( 1 - 20%) = 216

<=> X x 4/5 = 216

<=> X = 270

a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{0;3;-3\right\}\)

b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)

=>-6x=5

hay x=-5/6

Lời giải:
$7^{2x-6}=49=7^2$

$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

b) \(\left(x-1\right)^3=\dfrac{1}{8}\)

    \(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)

     \(x-1=\dfrac{1}{2}\)

     \(x=\dfrac{1}{2}+1\)

     \(x=\dfrac{3}{2}\)

a) \(6x^2-72x=0\)

\(6x\left(x-12\right)=0\)

\(6x=0\) hoặc \(x-72=0\)

*) \(6x=0\)

\(x=0\)

*) \(x-12=0\)

\(x=12\)

Vậy \(x=0;x=12\)

b) \(-2x^4+16x=0\)

\(-2x\left(x^3-8\right)=0\)

\(-2x=0\) hoặc \(x^3-8=0\)

*) \(-2x=0\)

\(x=0\)

*) \(x^3-8=0\)

\(x^3=8\)

\(x=2\)

Vậy \(x=0;x=2\)

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

\(x^2-5x-x^2+6x-9=0\)

\(x-9=0\)

\(x=9\)

d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)

\(x^3-6x^2+12x-8-x^3+8=0\)

\(-6x^2+12x=0\)

\(-6x\left(x-2\right)=0\)

\(-6x=0\) hoặc \(x-2=0\)

*) \(-6x=0\)

\(x=0\)

*) \(x-2=0\)

\(x=2\)

Vậy \(x=0;x=2\)