\(A=\frac{1}{21}+\frac{1}{2.22}+.......\frac{1}{1993.2013}\)
ai lam nhanh to tich cho
Những câu hỏi liên quan
Tính frac{A}{B}biết : A frac{1}{21}+frac{1}{2.22}+ ...............+ frac{1}{n.left(n+20right)}+........................+frac{1}{1993.2013} B frac{1}{1994}+ frac{1}{2.1995}+..................+frac{1}{n.left(n+1993right)}+...................+frac{1}{20.2013}
Đọc tiếp
Tính \(\frac{A}{B}\)biết :
A = \(\frac{1}{21}\)+\(\frac{1}{2.22}\)+ ...............+ \(\frac{1}{n.\left(n+20\right)}\)+........................+\(\frac{1}{1993.2013}\)
B = \(\frac{1}{1994}\)+ \(\frac{1}{2.1995}\)+..................+\(\frac{1}{n.\left(n+1993\right)}\)+...................+\(\frac{1}{20.2013}\)
Tính A/B biết :A= 1/21+1/2.22+...+ 1/n.(n+20)...+ 1/1993.2013, B= 1/1994+1/2.1995+...+ 1/n.(n+1993)+...+ 1/20.2013
GIÚP MIK VS NHA
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
CMR \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2+b^2+c^2}\)
Ai nhanh mk tich cho 3 cai
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+abc+a^2c+b^2a+b^2c+abc+c^2b+c^2a=0\)
\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(ab+ac+bc+c^2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
So ez
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Cho \(A=\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+......+\frac{1}{80.100}\)
\(B=\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+....+\frac{1}{20.100}\). Tính \(\frac{A}{B}\)
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
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\(\frac{a;a+5}{25}=\frac{1}{5}\)
ai nhanh tay minh tich 5 lai cho nhe !
bai 1 :thuc hien phep tinh
a)\(\frac{-5}{2}:\left(\frac{3}{4}-\frac{1}{2}\right)\) b) \(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\) c)\(\left(\frac{-5}{24}+0,75+\frac{7}{12}\right):\left(-2\frac{1}{4}\right)^2\)
AI LAM NHANH VA DUNG THI MK SE TICK CHO NHA
MK DAG CAN GAP !!!!!!! CAC BAN LAM NHANH LEN NHA
1) Tìm các cặp số nguyên (x,y) sao cho \(\frac{x}{16}-\frac{1}{y}=\frac{1}{32}\)
2) Cho A=\(\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
B=\(\frac{1}{1.81}+\frac{1}{2.82}+\frac{1}{3.83}+...+\frac{1}{20.100}\) Tính:\(\frac{A}{B}\)
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 \(20A=80B\)
\(\Rightarrow\frac{A}{B}=\frac{80}{20}=4\)
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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 \(\left(x;y\right)=\left(1;32\right);\left(0;-32\right)\)
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tinh gia try bieu thuc
A=\(\left[\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}\right]:\left[\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}\times\frac{1}{23}\times\frac{1}{1009}\right]+1:\left[30\times1009-160\right]\)
ai lam dung co ca cach lam cho 1 like
So sanh S=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{40}+\frac{1}{42}voi\frac{1}{2}\)
minh can cach lam
ai nhanh minh tick
Nhận xét: \(\frac{1}{5}< \frac{1}{42};\frac{1}{9}< \frac{1}{42};\frac{1}{10}< \frac{1}{42};\frac{1}{40}< \frac{1}{42}\)
\(\Rightarrow S< \frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}\)
\(\Rightarrow S< \frac{5}{42}< \frac{21}{42}=\frac{1}{2}\)
Vậy S < 1/2
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1/5+1/9+1/10+1/40+1/42=1159/2520
1159/2520=0.4599.....
1/2=0.5
Mà:0.5>0.4599
Nên:1/5+1/9+1/10+1/40+1/42>1/2
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