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Ta có
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\) nên
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)
\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)
Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự
\(A=\frac{1}{1.2}-x+\frac{1}{2.3}-x+...+\frac{1}{100.101}-x+100x\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}-100x+100x\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\) vậy
\(\frac{11}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}\div\frac{11}{45}=2\)
vậy suy ra x =2
mình chắc chắn 100% luôn đó, cái này ở trong violympic toán 7 vòng 12 phải ko
Mình chỉ làm cho bạn câu d và e thôi
d) ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +....... +1/99 - 1/100 ) . (x - 3)=1
( 1 - 1/100 ) . (x - 3 )=1
99/100.(x -3)=1
x - 3 = 1:99/100
x - 3 =100/99
x = 100/99 + 3
x = 397/99
e) (1/2 . (1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +.....+1/99 - 1/101 ) . (x+2) =3/101
(1/2 . ( 1 - 1/101 ).(x+2)=3/101
(1/2 . 100/101 ) . (x + 2) =3/101
100/202 . ( x + 2 )= 3/101
50/101 . (x + 2 ) = 3/101
x + 2 = 3/101 :50/101
x+2=3/50
x =3/50-2
x= -97/100
\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{x}{x.(x+1)}\) = \(\dfrac{44}{45}\)
\(\Rightarrow\) 1 - \(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{44}{45}\)
\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{44}{45}\)
\(\Rightarrow\) \(\dfrac{x}{x+1}\) = \(\dfrac{44}{45}\)
\(\Rightarrow\) \(x=44\)
<=> 1/2-1/3+1/3-1/4+...+1/x-1/x+1 = 44/45
<=> 1/2-1/x+1 = 44/45
<=> 1/x+1 = 1/2 - 44/45 = -43/90
=> x+1 = -90/43
=> x = -133/43
k mk nha