\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{18}{19}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{18}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
...............
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 nha
Thái Lâm Hoàng
Đặt \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)
1/3 . (1/1-1/4+1/4-1/7+1/7-1/10+...+1/x-(x+3)=6/19
1-1/x+3 =6/19.3=2/19
1/x+3=1-2/19=17/19
17.x+3 =19
x+3=19/17
x=19/17-3
x=-32/17
nếu đúng cho mình sao nha pạn
