x+3x+5x+...+17x+19x=100

x+3+5+...+17+19=100

x+99=100

x=100-99

x=1

Sansibailop5 ._.

X + 3X + 5X + ... + 17X + 19X = 100

<=> X x ( 1 + 3 + 5 + ... + 17 + 19 ) = 100

<=> \(X\times\frac{\left(19+1\right)\left[\left(19-1\right)\div2+1\right]}{2}=100\)

<=> X x 100 = 100

<=> X = 1

Không hiểu chỗ nào thì ib nhé :v 

\(\frac{2}{x+6}+\frac{1}{x+1}=\frac{3x+8}{x^2+7x+6}\left(ĐKXĐ:x\ne-6;-1\right)\)

\(\Leftrightarrow\frac{2\left(x+1\right)+x+6}{x^2+7x+6}=\frac{3x+8}{x^2+7x+6}\)

 \(\Leftrightarrow2x+2+x+6=3x+8\)

\(\Leftrightarrow3x+8=3x+8\)

          \(\Rightarrow\)x vô số nghiệm

3x + 5x + 1x = 27

x . (3+5+1) = 27

x. 9 = 27

x= 27:9 = 3

ĐÚNG GIÚP MK NHÉ

3x + 5x + x = 27

9x = 27

  x = 3

Vậy x = 3

\(3\times+5\times+\times=27\)

\(\Rightarrow\left(3+5+1\right)\cdot\times=27\)

\(\Rightarrow9\times=27\)

\(\Rightarrow\times=3\)

bằng 2

2xy +3x =2y +9

<=>2xy - 2 + 3x -3 = 9-3

<=>2y(x-1) +3(x-1) = 6

<=>(x-1)(2y+3) =6

ta có bảng: 

Vậy: \(\left(x,y\right)\in\) {(3,0);(-1,-3)...bạn tự liệt kê ra​}

có điều kiện của x,y không?

a)(3x+2):8=4

3x+2=4.8

3x+2=32

3x=32-2

3x=30

x=30:3

x=10

b)x+2x+3x+4x=100

x(1+2+3+4)=100

x.10=100

x=100:10

x=10

b) x + 2x + 3x + 4x = 100

x( 1 + 2 + 3 + 4 ) = 100

x . 10 = 100

x = 100 : 10

x = 10

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

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

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

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

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

\(\text{| 3x + 1 | = 4}\)

\(\text{3x = 3}\)

\(\text{x = 1}\)

\(\text{Trị tuyệt đối cũng bằng 1}\)