\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)

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

...............

đặt VT là A ta có:

\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{6}{19}\)

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

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

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

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

\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{6}{19}\)

\(\Rightarrow\frac{1}{3}\cdot\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{6}{19}\)

\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{6}{19}\cdot3=\frac{2}{19}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{2}{19}=\frac{17}{19}\)

=>17*(x+3)=19

x+3=19/17

x=19/17-3

x=-32/17

nếu thấy đúng thì k cho mh nhaThái Lâm Hoàng

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
 

\(C=2.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

 \(=2.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

  \(=2.\left(1-\frac{1}{100}\right)\)

 \(=2.\frac{99}{100}=\frac{198}{100}\)

C = \(3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

C = \(3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

C = 3 \(\left(1-\frac{1}{100}\right)\)

C = 3 \(\left(\frac{100}{100}-\frac{1}{100}\right)\)

C = \(3.\frac{99}{100}\)

C = \(\frac{297}{100}\)

phải đặt thừa số chung là 2 nha

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

    \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

      \(=1-\frac{1}{46}< 1\)

Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)

cả 2 cái cộng lại hay là từng cái một vậy bạn?

a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?

\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)

b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)

=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)

=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)

lp 6  lm bài lp 7 lm j

tí nữa lm cho đag mải

đề dễ mà định thi vao đâu vậy

\(A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\left(1-\frac{1}{100}\right)\)

\(A=\frac{297}{100}\)

Lo xa ha Lan Hương nhỉ kb nha mk cx lp 6 nè