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b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)
=>x+99/100=5/3
=>x=5/3-99/100=203/300
c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
=>5-x=7/3
hay x=8/3
\(\Rightarrow\frac{11}{12}+\frac{11}{12}-\frac{11}{23}+...+\frac{11}{89}-\frac{11}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+...+\frac{1}{89}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{1}{12}-\frac{1}{100}+x=\frac{2}{3}\)
\(\Rightarrow\frac{11}{150}+x=\frac{2}{3}\)
=>\(x=\frac{2}{3}-\frac{11}{150}\)
=>x=\(\frac{89}{150}\)
Ta có:
\(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)+x=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{99}{100}+x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{99}{100}=\dfrac{-97}{300}\)
Vậy \(x=\dfrac{-97}{300}\)
Ngày thứ ba đội đó sửa được số đoạn đường là:
1- (\(\frac{5}{9}+\frac{1}{4}\)) = \(\frac{7}{36}\)(đoạn đường)
Đoạn đường đội đó sửa được số mét là:
7 : 7/36= 36(mét
đáp số : 36 mét
câu 2:
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{2}{3}\) \(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\) \(1-\frac{1}{100}+x=\frac{2}{3}\) \(\frac{99}{100}+x=\frac{2}{3}\) \(x=\frac{2}{3}-\frac{99}{100}\) \(x=\frac{200}{300}-\frac{297}{300}\) \(x=\frac{-97}{300}\)
Bài giải
6 km của đoạn đường cần sửa dài là:
1 - 2/5 - 3/7 = 6/35 ( cả đoạn đường )
Độ dài đoạn đường đội đó đã sửa là:
6 : 6/35 = 35 ( km )
Đáp số : 35 km
a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)
\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)
\(\Rightarrow x=\dfrac{203}{300}\)
b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)
=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)
\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)
\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)
\(\Rightarrow x=\dfrac{515}{264}\)
c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(\frac{1}{1}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{23}\right)+\left(\frac{1}{23}-\frac{1}{34}\right)+...+\left(\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(1-\frac{1}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}\)
\(\Leftrightarrow\)\(x=\frac{203}{300}\)
Vậy \(x=\frac{203}{300}\)
\(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\left(1-\frac{1}{100}\right)+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(\frac{99}{100}+x=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\frac{5}{3}-\frac{99}{100}=\frac{203}{300}\)
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\frac{99}{100}-x=\frac{2}{3}\)
\(x=\frac{99}{100}-\frac{2}{3}\)
\(x=\frac{97}{300}\)
b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)
\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)
\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)
\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)
\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)
\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)
\(\Rightarrow x=\frac{97}{300}\)
b, k hiểu đề :v
a) ( 11 12 + 11 12.23 + 11 23.34 + . . . + 11 89.100 ) − x = 2 3 ( 12 11 + 12.23 11 + 23.34 11 +...+ 89.100 11 )−x= 3 2 ( 1 − 1 12 + 1 12 − 1 23 + 1 23 − 1 34 + . . . + 1 89 − 1 100 ) − x = 2 3 (1− 12 1 + 12 1 − 23 1 + 23 1 − 34 1 +...+ 89 1 − 100 1 )−x= 3 2 ( 1 − 1 100 ) − x = 2 3 (1− 100 1 )−x= 3 2 99 100 − x = 2 3 100 99 −x= 3 2 x = 99 100 − 2 3 x= 100 99 − 3 2 x = 97 300 x= 300 97 b) x + 1 9 + x + 3 7 + x + 5 5 + x + 7 3 = a 9 x+1 + 7 x+3 + 5 x+5 + 3 x+7 =a ⇒ x + 1 9 + 1 + x + 3 7 + 1 + x + 5 5 + 1 + x + 7 3 + 1 = a + 4 ⇒ 9 x+1 +1+ 7 x+3 +1+ 5 x+5 +1+ 3 x+7 +1=a+4 x + 10 9 + x + 10 7 + x + 10 5 + x + 10 3 = a + 4 9 x+10 + 7 x+10 + 5 x+10 + 3 x+10 =a+4 ( x + 10 ) . ( 1 9 + 1 7 + 1 5 + 1 3 ) = a + 4 (x+10).( 9 1 + 7 1 + 5 1 + 3 1 )=a+4