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

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

mở dấu trị tuyệt đối ra rồi tính như bình thường

x=24( đang nghĩ cách làm )

\(1+2+3+...+x=300\)

\(\Rightarrow\frac{x.\left(x+1\right)}{2}=300\)

\(\Rightarrow x.\left(x+1\right)=600\)

\(\Rightarrow x.\left(x+1\right)=24.25\)

\(\Rightarrow x=24\)

PP/ss: Hoq chắc ạ_333

\(12,45\times x-0,45\times x+x=3,9.\)

\(\Leftrightarrow x\times\left(12,45-0,45+1\right)=3,9.\)

\(\Leftrightarrow x\times13=3,9.\)

\(\Leftrightarrow x=0,3.\)

Vậy \(x=0,3.\)

b) 200 x 4 = .800.....              300 x 2 = ..600....

200 x 2 = .400.....             300 x 3 = ...900...

400 x 2 = ...800...             500 x 1 = ..500....

100 x 4 = ..400....             100 x 3 = .300.....

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x=2

x=500

câu 1 x = 2

câu 2 x = 500