1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19

1/3.(1-1/x+3)=6/19

1-1/x+3=6/19:1/3

1-1/x+3=18/19

1/x+3=1-18/19

1/x+3=1/19

=> x+3=19

=>x=19-3

x=16    

Đặt biểu thức là A, ta có:

3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)

3A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)

3A=1-\(\frac{1}{x+3}\)

A=\(\frac{1}{3}-\frac{3}{x+3}\)

=>\(\frac{1}{3}-\frac{3}{x+3}\)  =\(\frac{6}{19}\) =>x=168

1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19

1/3.(1-1/x+3)=6/19

1-1/x+3=6/19:1/3

1-1/x+3=18/19

1/x+3=1-18/19

1/x+3=1/19

=> x+3=19

=>x=19-3

x=16    

đặt VT là A ta đc:

\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)

\(3A=1-\frac{1}{x+3}\)

\(A=\left(1-\frac{1}{x+3}\right):3\)

thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

=>x+3=19

=>x=16
 

bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được

Ta có:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)

\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)

\(1-\frac{1}{x+3}=\frac{375}{376}\)

\(\frac{x+2}{x+3}=\frac{375}{376}\)

=> x + 2 = 375

=> x = 375 - 2

=> x = 373

Pk bt tổng này bằng bao nhiêu ms tính đc chứ

3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)

3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)

1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3

1/1 - 1/x+3

x+3/x+3 - 1/x+3

x+2/x+3

Đặt \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\)

\(3A=\frac{3}{1.4}+\frac{4}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\)

\(3A=1-\frac{1}{x+3}\)

\(A=\frac{1-\frac{1}{x+3}}{3}\)

Cái này mình rút  gọn rồi đó

\(\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)=\frac{0,33x}{2009}\)

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)

\(\left(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

\(1-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{2009}\Rightarrow2009x99=0,33x\times100\)

198891:100:0,33=6027=x