a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555

=> 101x +5050 = 5555

=> 101x = 505

=> x = 505 : 101 = 5

Vậy, x = 5

b)1+2+3+4+...+x=820

=> ( x+1) x :2 = 820

=> (x+1)x = 1640

Mà 1640 = 40 . 41

=> x = 40 ( vì {x+1} - x = 1)

Vậy, x = 40

c) 3^x+1 = 9.27=243

=> 3^x+1 = 3⁵

=>x + 1 = 5

=> x = 4

Vậy, x=4

d) x+2x+3x+...+99x+100x=15150

=> [( 100 + 1) x 100 :2 ] x = 15150

=> 5050x = 15150

=> x = 15150:5050 = 3

Vậy, x =3

e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550

=> 100x + 5050 = 205550

=> 100x =  205550 - 5050= 200500

=> x =  200500 : 100 = 2005

Vậy, x = 2005

f)3^x+3^x+1+3^x+2=351

=> 3^x + 3^x . 3 + 3^x x 9 = 351

=> 3^x ( 1+3+9) = 351

=> 3^x  . 13 = 351

=> 3^x  = 351 :13=27 mà 27 = 3³

=> x=3

Vậy, x=3

mình đg cần gấp á

a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)

\(\Rightarrow101\cdot x+5050=5555\)

\(\Rightarrow101\cdot x=5555-5050\)

\(\Rightarrow101\cdot x=505\)

\(\Rightarrow x=505:101\)

\(\Rightarrow x=5\)

b) \(1+2+3+4+...+x=820\)

\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)

\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)

\(\Rightarrow\left(x+1\right)\cdot x:2=820\)

\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=1640\)

Ta thấy: \(40\cdot41=1640\)

Vậy: \(x=40\)

đặt 3^x ra ngoài bạn nhé bàn phím mik hỏng rồi ;-;

$\Rightarrow 3^x(1+3+3^2+3^3)=1080$

$\Rightarrow 3^x.40=1080$

$\Rightarrow 3^x=27=3^3$

$\Rightarrow x=3$

\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=1080\)

\(\Rightarrow3^x+3^x.3^1+3^x.3^2+3^x.3^3=1080\)

\(3^x.\left(1+3+9+27\right)=1080\)

\(3^x.40=1080\)

\(3^x=27\)

\(3^x=3^3\)

Vậy: x = 3

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

\(\Leftrightarrow9x^2-3x-9x^2+1=0\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)

b: Ta có: \(x^2-5x+25-5x=0\)

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

\(\Leftrightarrow x-5=0\)

hay x=5

\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=9x^2+24x+16-9x^2+1=24x+17\)

Đặt \(24x+17=0\Leftrightarrow x=-\frac{17}{24}\)

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

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

a: Bạn ghi lại đề đi bạn

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

\(3x+3x+1+3x+2=117\)

\(\Rightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)

\(\Rightarrow9x+3=117\)

\(\Rightarrow9x=117-3\)

\(\Rightarrow9x=114\)

\(\Rightarrow x=114:9\)

\(\Rightarrow x=\frac{38}{3}\)

Vậy \(x=\frac{38}{3}\)

P/s : Đúng nha 

~ Ủng hộ nhé 

=> 6x + 3= 117

6x = 117-3

6x= 114

x=19

=>9x +3 =117

9x=114

x=38/3

x²+3x²+3x+1-3x²-3x = 0

  => x³+1 = 0

  => x³     = 0-1

  => x³     = -1

  => x       = -1

\(x^3+3x^2+3x+1-3x^2-3x=0\)0

\(\Leftrightarrow x^3+\left(3x^2-3x^2\right)+\left(3x-3x\right)+1=0\)

\(\Leftrightarrow x^3+1=0\)

\(\Leftrightarrow x^3=1\)

\(\Leftrightarrow x^3=1^3\)

\(\Rightarrow x=1\)

lớp 8 đến bây giờ đã học hằng đẳng thức chưa ạ ? Để mình đưa bạn hướng giải câu đầu ! 

câu sau : x^2 -9^2 = 0<=> (x-9)^2=0 <=>x-9=0 <=> x=9

X^2-81=0

=>X^2-9^2=0

=>(X-9)^2=0

=>X-9=0

=>X=9