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5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)
3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)
1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3
1/1 - 1/x+3
x+3/x+3 - 1/x+3
x+2/x+3
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1
(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)
(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)
\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)
(44\(x\) + 3) : \(\frac{11}{5}\) = - 4
44\(x\) + 3 = - 4 x \(\frac{11}{5}\)
44\(x\) + 3 = - \(\frac{44}{5}\)
44\(x\) = - \(\frac{44}{5}\) - 3
44\(x\) = - \(\frac{59}{5}\)
\(x=-\frac{59}{5}:44\)
\(x\) = - \(\frac{59}{250}\)
Vậy \(x\) = - \(\frac{59}{250}\)
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
\(S=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{2002\cdot2005}\)
\(3S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2002\cdot2005}\)
\(3S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
\(3S=\frac{1}{1}-\frac{1}{2005}\)
\(3S=\frac{2004}{2005}\)
\(S=\frac{2004}{2005}\div3=\frac{668}{2005}\)
Ta có:
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2002.2005}\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2002.2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2002}-\frac{1}{2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{2005}\right)=\frac{1}{3}.\frac{2004}{2005}=\frac{668}{2005}\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(=\frac{1}{3}.\left(1-\frac{1}{100}\right)=\frac{x}{2}\)
\(\frac{1}{3}.\frac{99}{100}=\frac{x}{2}\)
\(\frac{99}{300}=\frac{x}{2}\)
\(x\)ko thỏa mãn
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.9}+......+\frac{1}{97.100}=\frac{x}{2}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{100}\right)=\frac{x}{2}\Rightarrow\frac{1}{3}.\frac{99}{100}=\frac{x}{2}\Rightarrow\frac{33}{100}=\frac{x}{2}\Rightarrow\frac{33}{100}=\frac{50x}{100}\Rightarrow33=50x\Rightarrow x=\frac{33}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)
bài 2 tính trong ngoặc tương tự bài trên rồi tìm x
bài 3
vì giá trị nguyên của x để B là 1 số nguyên
\(\Rightarrow x+4⋮x+3\)
lập bảng
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{197.200}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{197}-\frac{1}{200}\)
\(=1-\frac{1}{200}\)
\(=\frac{199}{200}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{197}-\frac{1}{200}\)
\(A=1-\frac{1}{200}\)
\(A=\frac{199}{200}\)
\(\frac{x}{1.4}+\frac{x}{4.7}+\frac{x}{7.10}+...+\frac{x}{36.39}=1\)
\(\frac{x}{3}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{36.39}\right)=1\)
\(\frac{x}{3}.[(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+...+(\frac{1}{36}-\frac{1}{39})]=1\)
\(\frac{x}{3}.(\frac{1}{1}-\frac{1}{39})=1\)
\(\frac{x}{3}.\frac{38}{39}=1\)
\(\frac{x}{3}=1:\frac{38}{39}\)
\(\frac{x}{3}=\frac{39}{38}\)
\(\Rightarrow x=.....\)
Mình tính vội nên không tính kết quả đúng chưa, cậu kiểm tra lại nha, còn cách làm thế là chuẩn rồi! Học tốt!