\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)

\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+3\right)}=\frac{100}{101}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)

\(\Rightarrow x+3=101\)

\(=>x=98\)

                               \(D=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)

                                \(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)

                               \(D=1-\frac{1}{x+3}=\frac{100}{101}\)

                            \(D=\frac{1}{x+3}=1-\frac{100}{101}\)

                             \(D=\frac{1}{x+3}=\frac{1}{101}\)

                         \(\Rightarrow x+3=101\Rightarrow x=98\)

                      Ủng hộ mk nha ^_^

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
 

a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1  - \(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|

\(\Rightarrow\)\(\frac{x}{3}\)= \(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)

Với \(\frac{x}{3}\) = \(\frac{33}{100}\)

\(\Rightarrow\)100x= 33.3

 \(\Rightarrow\)100x=99

\(\Rightarrow\)x=\(\frac{99}{100}\)

Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)

\(\Rightarrow\)100x=-33.3

\(\Rightarrow\)100x=-99

\(\Rightarrow\)x=\(\frac{-99}{100}\)

Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)

b, \(\frac{4}{1.5}\)+ \(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)

\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|

\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)

Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)

\(\Rightarrow\)(5x-4).101=100.101

\(\Rightarrow\)505x-404=10100

\(\Rightarrow\)505x=10504

\(\Rightarrow\)x=\(\frac{104}{5}\)

Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)

\(\Rightarrow\)(5x-4). 101=-100.101

\(\Rightarrow\)505x-404=-10100

\(\Rightarrow\)505x=-9696

\(\Rightarrow\)x=\(\frac{-96}{5}\)

Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)

a) (x-1)+(x-2)+(x-3)+...+(-100)=101

(x+x+x+...+x)-(1+2+3+...+100)=101

=> 100x-5050=101

100x=101+5050

100x=5151

x=5151:100

x=5151/100

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

Đặt vế trái phương trình là A

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

\(3A=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{\left(x+3\right)-x}{x\left(x+3\right)}\)

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

\(3A=1-\frac{1}{x+3}=\frac{x+2}{x+3}\Rightarrow A=\frac{x+2}{3\left(x+3\right)}\)

\(\Rightarrow\frac{x+2}{3\left(x+3\right)}=\frac{667}{2002}\Rightarrow2002\left(x+2\right)=3.667.\left(x+3\right)\)

\(\Leftrightarrow2002x+4004=2001x+6003\Leftrightarrow x=1999\)

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

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

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

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

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