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\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}...\frac{1-98^2}{98^2}.\frac{1-99^2}{99^2}\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{98^2-1}{98^2}.\frac{99^2-1}{99^2}\)
= \(\frac{\left(2-1\right).\left(2+1\right)}{2^2}.\frac{\left(3-1\right).\left(3+1\right)}{3^2}.\frac{\left(4-1\right).\left(4+1\right)}{4^2}...\frac{\left(98-1\right)\left(98+1\right)}{98^2}.\frac{\left(99-1\right)\left(99+1\right)}{99^2}\)
\(=\frac{\left(2-1\right).\left(3-1\right).\left(4-1\right)...\left(99-1\right)}{2.3.4...98.99}.\frac{\left(2+1\right).\left(3+1\right).\left(4+1\right)...\left(99+1\right)}{2.3.4...98.99}\)
\(=\frac{1.2.3....98}{2.3.4...98.99}.\frac{3.4.5...100}{2.3.4...98.99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{50}{99}\)
Chúc bạn học tốt !!!
Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+...+\dfrac{x+50}{50}+1=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}\right)=0\)
mà \(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}>0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+...+\left(\dfrac{x+50}{50}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}\right)=0\)
\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}>0\) )
\(\Leftrightarrow x=-100\)
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 ~
Cộng 1 vào từng phân số ta sẽ đc
\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
\(\Rightarrow x=-100\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\)
<=> \(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
<=> x + 100 = 0 (vì \(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)\ne0\))
<=> x = -100
⇔ \(\frac{a+1}{99}+1+\frac{a+2}{98}+1=\frac{a+3}{97}+1+\frac{a+4}{96}+1\)
⇔ \(\frac{a+100}{99}+\frac{a+100}{98}=\frac{a+100}{97}+\frac{a+100}{96}\)
⇔ (a+100)(\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\)) = 0
mà \(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\) ≠ 0
⇒ a+100 = 0
⇒ a = -100
Vậy a=-100
=1/100+1/99-1/99+1/98-1/98+1/97-...........-1/2+1/2-1/2+1
=1/100+1
=101/100
Soái ca 2k6 Làm đi bạn !!
\(\frac{3^{2}+1}{3^{2}-1}+\frac{5^{2}+1}{5^{2}-1}+...+\frac{99^{2}+1}{99^{2}-1}=49+\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}=49+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}=49.49\)