(1+1/2) x (1+1/3) x (1+1/4) x ......x(1+1/98) x (1+1/99)
Ai nhanh mik tik (có cách làm nhanh)
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\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times.....\times1\frac{1}{98}\times1\frac{1}{99}.\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times.....\times\frac{99}{98}\times\frac{100}{99}\)
= \(\frac{3\times4\times5\times....\times99\times100}{2\times3\times4\times.....\times98\times99}\)
= \(\frac{100}{2}\)
= \(50\)
\(\left(x-2\right)\left(x-5\right)< 0\)
Xét các trường hợp :
Trường hợp 1:
\(\hept{\begin{cases}x-2< 0\\x-5>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 2\\x>5\end{cases}}}\Rightarrow\varnothing\)
Trường hợp 2:
\(\hept{\begin{cases}x-2>0\\x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>2\\x< 5\end{cases}\Rightarrow2< x< 5}\)
Vậy \(2< x< 5\)thì\(\left(x-2\right)\left(x-5\right)< 0\)
\(\frac{x-1}{2}=\frac{x}{4}\)
\(\Leftrightarrow\)\(\frac{2\left(x-1\right)}{4}=\frac{x}{4}\)
\(\Leftrightarrow\)\(2\left(x-1\right)=x\)
\(\Leftrightarrow\)\(2x-2=x\)
\(\Leftrightarrow\)\(2x-x=2\)
\(\Leftrightarrow\)\(x=2\)
Vậy....
\(\frac{x+1}{x-1}=\frac{1}{2}\)
\(\Leftrightarrow\)\(2\left(x+1\right)=x-1\)
\(\Leftrightarrow\)\(2x+2=x-1\)
\(\Leftrightarrow\)\(2x-x=-1-2\)
\(\Leftrightarrow\)\(x=-3\)
Vậy....
a) \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{99.101}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)+\left(\frac{1}{2.4}+...+\frac{1}{98.100}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)+2.\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{101}\right)+2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=2\cdot\frac{100}{101}+2\cdot\frac{49}{100}=\frac{200}{101}+\frac{49}{50}\)
câu b mk ko bk! xl bn nha!
mk nhầm
...
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\) 1/100)
= 1/2.(1-1/101) + 1/2.(1/2-1/100)
=1/2.100/101 + 1/2.49/100
= 50/101 + 49/200
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{100}{2}=50\)
Thanks nhìu