tìm x:
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
Bài 10: Phép nhân phân số
Ta có:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3.....9}{2.3.4.....10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
\(\Leftrightarrow x=201\)
Vậy x = 201
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\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{10}\right)=\)\(\dfrac{x}{2010}\)
\(\Leftrightarrow\) \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}......\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3......9}{2.3.4.......10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow2010=10x\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
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tìm x
(x+1)+\(\left(x+\dfrac{1}{3}\right)+\left(x+\dfrac{1}{6}\right)+...+\left(x-\dfrac{1}{55}\right)=11\dfrac{9}{11}\)
\(\dfrac{2}{3}\cdot x-70\dfrac{10}{11}:\left(\dfrac{13}{15}+\dfrac{1313}{1515}+\dfrac{13}{63}+\dfrac{1313}{6363}\right)\)
Chứng tỏ
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Đặt A= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)
= \(\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\right)\)
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\(\dfrac{6}{1\cdot3\cdot7}+\dfrac{6}{6\cdot7\cdot9}+...+\dfrac{6}{13\cdot15\cdot19}\)
Giúp mình nha
\(\dfrac{6}{1\cdot3\cdot5}+\dfrac{6}{5\cdot7\cdot9}+...+\dfrac{6}{13\cdot15\cdot17}\)
\(\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}-...-\dfrac{92}{100}}{\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}}\)
Tính một cách hợp lý:
aleft(dfrac{1}{2}-1right)left(dfrac{1}{3}-1right)...left(dfrac{1}{100}-1right)) x:dfrac{99}{100}:dfrac{98}{99}:...:dfrac{2}{3}:dfrac{1}{2}
b) dfrac{5-dfrac{5}{3}+dfrac{5}{9}-dfrac{5}{27}}{8-dfrac{8}{3}+dfrac{8}{9}-dfrac{8}{27}}:dfrac{15-dfrac{15}{11}+dfrac{15}{121}}{16-dfrac{16}{11}+dfrac{16}{121}}
c) dfrac{dfrac{1}{9}-dfrac{5}{6}-4}{dfrac{7}{12}-dfrac{1}{36}-10}
d) left(dfrac{1}{2}+1right)left(dfrac{1}{3}+1right)...left(dfrac{1}{99}+1right)
e)
Đọc tiếp
Tính một cách hợp lý:
a\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)\)) \(x:\dfrac{99}{100}:\dfrac{98}{99}:...:\dfrac{2}{3}:\dfrac{1}{2}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
c) \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
d) \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{99}+1\right)\)
e)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
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5/6+6 và 5/6.(11 và 5/20-9 và 1/4):8 và 1/3
dựa vào lược đồ tự nhiên Trung và Nam Mĩ nêu sự phân bố của một số thảm thực vật như Rừng rậm xanh quanh năm;rừng thưa Xa-van;thảo nguyên;hoang mạc và bán hoang mạc
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-15/12 . x+ 3/7=6/5 .x -1/2
\(-\dfrac{15}{12}x+\dfrac{3}{7}=\dfrac{6}{5}x-\dfrac{1}{2}\)<=>\(-\dfrac{5}{4}x+\dfrac{3}{7}=\dfrac{6}{5}x-\dfrac{1}{2}\)
<=>\(\dfrac{3}{7}+\dfrac{1}{2}=\dfrac{6}{5}x+\dfrac{5}{4}x\)<=>\(\dfrac{3}{7}+\dfrac{1}{2}=\left(\dfrac{6}{5}+\dfrac{5}{4}\right)x\)
<=>\(\dfrac{3.2+7}{2.7}=\dfrac{\left(4.6+5.5\right)x}{4.5}\Leftrightarrow x=\dfrac{\left(3.2+7\right).4.5}{2.7\left(4.6+5.5\right)}\)
\(x=\dfrac{\left(3.2+7\right).2.5}{7\left(4.6+5.5\right)}=\dfrac{13.10}{7^3}=\dfrac{130}{243}\)
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