\(VT>0\Rightarrow VP>0\Rightarrow x>0\)
Xóa dấu giá trị tuyệt đối là : \(\left(397-1\right):4+1=100\)
\(\Rightarrow100x=\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\right)\)
\(\Rightarrow x=\left(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\right)\)
\(\Rightarrow4x=1-\frac{1}{401}\)
\(\Rightarrow x=\frac{100}{401}\) ( thõa mãn )
Chứng tỏ : \(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}< 1\)
Gíup mình nha
Tìm x
\(\dfrac{x}{5}=\dfrac{x+6}{15}\)
Tính tổng S
\(S=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{17.21}\)
tìm x, biết:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
Tìm x:
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+.......+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
BT3: Tìm x, biết
21) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
1 phan 5 nhan 8 cong 1 phan 8 nhan 11 cong 1 phan 11 nhan 14 +..........+1 phan x nhan (x+3)=101 phan 1540
1 phan 5 nhan 8 cong 1 phan 8 nhan 11 cong 1 phan 11 nhan 14 +..........+1 phan x nhan (x+30+101 phan 1540
Tìm \(x\) biết :
\(\frac{1}{5.8}.\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
Tìm x:
a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
