Ta có: \(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\)

\(=3\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{299\cdot300}\right)\)

\(=3\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{299}-\frac{1}{300}\right)\)

\(=3\cdot\left\lbrack1+\frac12+\frac13+\frac14+\cdots+\frac{1}{299}+\frac{1}{300}-2\left(\frac12+\frac14+\cdots+\frac{1}{300}\right)\right\rbrack\)

\(=3\left(1+\frac12+\frac13+\cdots+\frac{1}{300}-1-\frac12-\cdots-\frac{1}{150}\right)\)

\(=3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

Ta có: \(\left(\frac{3}{1\cdot2}+\frac{3}{3\cdot4}+\cdots+\frac{3}{299\cdot300}\right)\cdot\left(x+\frac23\right)=\frac{2}{151}+\frac{2}{152}+\cdots+\frac{2}{300}\)

=>\(3\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\left(x+\frac23\right)=2\left(\frac{1}{151}+\frac{1}{152}+\cdots+\frac{1}{300}\right)\)

=>\(x+\frac23=\frac23\)

=>x=0

mở dấu trị tuyệt đối ra rồi tính như bình thường

Giải :

a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 50 )  = 2040

=> 51x + ( 1 + 2 + 3 + ... + 50 ) = 2040

=> 51x + ( 50 + 1 ) . 50 : 2 = 2040

=> 51x + 1275 = 2040

=> 51x = 765

=> x = 15

b) 1 + 2 + 3 + ... + x = 300

=> ( x + 1 ) . x : 2 = 300

=> ( x + 1 ) . x = 600

=> ( x + 1 ) . x = 25 . 24

=> x = 24

a) x = 44.             

b) x = 40.            

c) x = 95.             

d) x = 10.

a) x - 120: 30 = 40

    x    -40        =40

              x        =40+40

              x        =80

b) (x + 120) : 20 = 8

    (x+ 120)         = 8x20

     x+120           =160

     x                   = 160-120

     x                  = 40

c) (x + 5). 3 = 300

     x+5=300:3

     x+5=100

     x=100-5

     x=95

d) x.2 + 21 : 3= 27

    x.2   +7=27

      x.2   = 27-7

    x.2= 20

      x=20:2

      x=10

Đáp án cần chọn là: B

x=100

thử lại: (300-100)-100=200-100=100

x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135.

Hảo hán