(1-\(\frac{1}{97}\))x(1-\(\frac{1}{98}\))x...x(1-\(\frac{1}{1000}\))=?
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\(\left(1-\frac{1}{97}\right)x\left(1-\frac{1}{98}\right)x...x\left(1-\frac{1}{1000}\right)\)
\(\frac{96}{97}\cdot x\cdot\frac{97}{98}\cdot x\cdot...\cdot x\cdot\frac{999}{1000}\)
\(\frac{96}{97}\cdot\frac{97}{98}\cdot...\cdot\frac{999}{1000}\cdot x^{903}\)
\(\frac{96}{1000}\cdot x^{903}\)
\(\frac{12}{125}\cdot x^{903}\)
(1-1/97),(1-1/98).....(1-1/1000)
=96/97.97/98.....999/1000
=(96.97...999)/(97.98...1000)
=96/1000=12/125
Dấu . Là dấu nhân nhé
=\(\frac{96}{97}\)x\(\frac{97}{98}\)x...x\(\frac{999}{1000}\)
triệt tiêu đi ta có:
=\(\frac{96}{1000}\)=\(\frac{12}{125}\)
tick nhé!!
Ta có :
\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}=0\)
\(\Leftrightarrow\)\(\frac{x-100}{99}-\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}-\frac{x-100}{95}-\frac{x-100}{94}=0\)
\(\Leftrightarrow\)\(\left(x-100\right)\left(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)
Vì \(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)
Nên \(x-100=0\)
\(\Rightarrow\)\(x=100\)
Vậy \(x=100\)
Bài làm mang tính chất tham khảo vì em mới lớp 7 ~
\(\frac{x+1}{97}\) + \(\frac{x+1}{98}\) - \(\frac{x+1}{99}\) - \(\frac{x+1}{100}\) \(\Leftrightarrow\) (x+1).(1/97 + 1/98 - 1/99 - 1/100) . Vì (1/97 = 1/ 98 - 1/99 - 1/100) \(\ne\) 0 \(\Rightarrow\) x+ 1= 0 \(\Leftrightarrow\) x= -1
Ta có \(\left(1-\frac{1}{97}\right)\times\left(1-\frac{1}{98}\right)\times.....\times\left(1-\frac{1}{1000}\right).\)
\(=\frac{97-1}{97}\times\frac{98-1}{98}\times.....\times\frac{1000-1}{1000}\)
\(=\frac{96}{97}\times\frac{97}{98}\times....\times\frac{999}{1000}\) (rút gọn hết )
\(=\frac{96}{1000}\)
\(=\frac{12}{125}\)
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.........1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}......\frac{100}{99}\)
\(=\frac{\left(2.2\right).\left(3.3\right).\left(4.4\right).\left(5.5\right)....\left(10.10\right)}{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right).....\left(9.11\right)}\)
\(=\frac{\left(2.3.4.5...10\right).\left(23.4.5....10\right)}{\left(1.2.3.4...9\right).\left(3.4.5.6....11\right)}=\frac{10}{1}.\frac{2}{11}=\frac{20}{11}\)
b) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)
=\(\frac{96}{1000}\)