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Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
a: Số số hạng từ 0 đến 2012 là: 2012-0+1=2013(số)
Tổng là \(\dfrac{2012\cdot2013}{2}=2025078\)
Theo đề, ta có: 2013x+2025078=2033130
=>2013x=8052
hay x=4
b: \(\Leftrightarrow5\cdot3^{2x}+4\cdot3^{2x}\cdot3=1377\)
\(\Leftrightarrow3^{2x}\cdot17=1377\)
=>32x=81
=>x=2
\(VT>0\Rightarrow VP>0\Rightarrow x>0\)
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 )
\(\frac{8}{1.5}+\frac{8}{5.9}+\frac{8}{9.13}+...+\frac{8}{x\left(x+4\right)}=\frac{1}{2}\)
\(\Leftrightarrow\)\(2\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{1}{2}\)
\(\Leftrightarrow\)\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}=\frac{1}{4}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+4}=\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{x+4-1}{x+4}=\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{x+3}{x+4}=\frac{1}{2}\)
\(\Rightarrow\)\(2\left(x+3\right)=x+4\)
\(\Leftrightarrow\)\(2x+6=x+4\)
\(\Leftrightarrow\)\(x=-2\)
Vậy....
P/s: tham khảo mk ko chắc là đúng
a,\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
b,\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+....+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}=1=\frac{2008}{2008}\)
=> x = 2008
1, Tính tổng:
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)
2, Tìm x:
\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)
- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
`2.x-49=5.9`
`=> 2.x-49=45`
`=>2.x=45+49`
`=>2.x=94`
`=>x=94:2`
`=>x=47`
2 . x - 49 = 45
2 . x = 49 + 45
2 . x = 94
x = 94 : 2
x = 47