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Câu 2:
\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{3.23}+...+\frac{20}{80.100}\)
\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)
\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\) (1)
Lại có:
\(B=\frac{1}{1.81}+\frac{1}{2.82}+...+\frac{1}{20.100}\)
\(\Rightarrow80B=\frac{80}{1.81}+\frac{80}{2.82}+...+\frac{80}{20.100}\)
\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)
\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(2)
Từ (1) và (2) suy ra \(20A=80B\)
\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)
Câu 1:
\(\frac{x}{16}-\frac{1}{y}=\frac{1}{32}\)
\(\Leftrightarrow\frac{xy-16}{16y}=\frac{1}{32}\)
\(\Leftrightarrow\frac{xy-16}{y}=\frac{1}{2}\)
\(\Leftrightarrow2xy-32=y\)
\(\Leftrightarrow\left(2x-1\right).y=32\)
Tới đây ta nhận xét do \(2x-1\) luôn lẻ với mọi x nguyên nên \(2x-1\) là ước lẻ của 32
\(\Rightarrow2x-1=\left\{1;-1\right\}\)
Vậy: \(\left\{{}\begin{matrix}2x-1=1\\y=32\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x-1=-1\\y=-32\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-32\end{matrix}\right.\)
Có 2 cặp số nguyên thỏa mãn là \(\left(x;y\right)=\left(1;32\right);\left(0;-32\right)\)
20A=20/1.21+20/2.22+...+20/80.100
=1-1/21+1/2-1/22+...+1/80-1/100
=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)
80B=80/1.81+80/2.82+...+8/20.100
=1-1/81+1/2-1/82+...+1/20-1/100
=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)
=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)
=>20A=80B
=>A=4B
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
ta có: \(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{2.23}+...+\frac{20}{80.100}\)
\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)
\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{100}\right)\)
\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}+...+\frac{1}{100}\right)\)
lại có: \(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\)
\(80B=\frac{80}{1.81}+\frac{80}{2.82}+\frac{80}{3.83}+...+\frac{80}{20.100}\)
\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+\frac{1}{3}-\frac{1}{83}+...+\frac{1}{20}-\frac{1}{100}\)
\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}+...+\frac{1}{100}\right)\)
Vậy 20A = 80B
=> \(\frac{A}{B}=\frac{80}{20}=4\)
\(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
\(20A=\frac{20}{1.21}+\frac{20}{2.22}+\frac{20}{3.23}+...+\frac{20}{80.100}\)
\(20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+\frac{1}{3}-\frac{1}{23}+...+\frac{1}{80}-\frac{1}{100}\)
\(20A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{80}-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(20A=1+\frac{1}{2}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(1)
Lại có :
\(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\)
\(\Rightarrow80B=\frac{80}{1.81}+\frac{80}{2.82}+...+\frac{80}{20.100}\)
\(80B=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)
\(80B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)(2)
Từ (1) và (2) , suy ra : \(20A=80B\)
\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)
Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
bài 1 A là số nguyên <=> 3 chia hết cho (x-1) <=> (x-1) thuộc Ư(3) = { 1;-1;3;-3}
<=> x thuộc {2;0;4;-2}
3. + \(20A=\frac{21-1}{1\cdot21}+\frac{22-2}{2\cdot22}+...+\frac{100-80}{80\cdot100}\)
\(\Rightarrow20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+...+\frac{1}{80}-\frac{1}{100}\)
\(\Rightarrow20A=\left(1+\frac{1}{2}+...+\frac{1}{80}\right)-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{20}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
+ \(80B=\frac{81-1}{1\cdot81}+\frac{82-2}{2\cdot82}+...+\frac{100-2}{20\cdot100}\)
\(=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
\(\Rightarrow B=\frac{1}{80}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
Do đó : \(\frac{A}{B}=\frac{\frac{1}{20}}{\frac{1}{80}}=4\)
4. + \(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}=\frac{100}{2}=50\)