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\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)
\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x=100
Chúc bạn học tốt
( 100- 99+ 98- 97+ 96- 95+...+ 4- 3+ 2- 1): 50+ 2010- 12. X= 91.
A: 50+ 2010- 12. X= 91.
Xét A= 100- 99+ 98- 97+ 96- 95+...+ 4- 3+ 2- 1.
Dãy A có số các số hạng là:
( 100- 1): 1+ 1= 100( số)
Ta ghép 2 số hạng vào 1 nhóm được 50 nhóm.
A=( 100- 99)+( 98- 97)+( 96- 95)+...+( 4- 3)+( 2- 1).
A= 1+ 1+ 1+...+ 1+ 1( 50 số 1)
A= 1x 50.
A= 50.
=> A: 50+ 2010- 12. X= 91.
=> 50: 50+ 2010- 12. X= 91.
=> 1+ 2010- 12. X= 91.
=> 2011- 12. X= 91.
=> 12. X= 2011- 91.
=> 12. X= 1920.
=> X= 1920: 12.
=> X= 160.
Vậy X= 160.
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\left(\frac{x+1}{99}+\frac{99}{99}\right)+\left(\frac{x+3}{97}+\frac{97}{97}\right)+\left(\frac{x+5}{95}+\frac{95}{95}\right)=\left(\frac{x+2}{98}+\frac{98}{98}\right)+\left(\frac{x+4}{96}+\frac{96}{96}\right)+\left(\frac{\left(x+6\right)}{94}+\frac{94}{94}\right)\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{92}+\frac{x+100}{94}+\frac{x+100}{96}\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{92}-\frac{x+100}{94}-\frac{x+100}{96}=0\)
\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\right)=0\)
\(Mà\) \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\ne0\)
Nên x+ 100 = 0
x = 0 - 100 = -100
Vậy x= -100
cộng 1 vào mỗi tỉ số,ta được:
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)\(\Rightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+2+98}{98}+\frac{x+4+96}{96}+\frac{x+6+94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{98}+\frac{x+100}{96}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{98}-\frac{x+100}{96}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\ne0\)
=>x+100=0
=>x=-100
Vậy x=-100
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)
\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)
Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a
b)Tận cùng=5 hoặc 0 nhưng mình ngại viết lắm,thông cảm nha
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\Rightarrow\frac{x+3}{97}+1+\frac{x+5}{95}+1+\frac{x+4}{96}+1+\frac{x+1}{99}+1=0\)
\(\Rightarrow\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
Vậy x = -100