32. 22 - 32 . 19
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
12=9+19+2=42
28+12+22+18=80
79+1+71+9=160
32+22+32+22=108
Chúc bn hok giỏi
ta có: \(A=\frac{75}{100}+\frac{78}{21}+\frac{19}{32}+\frac{22}{21}+\frac{13}{32}\)
\(A=\frac{3}{4}+\left(\frac{78}{21}+\frac{22}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(A=\frac{3}{4}+4+\frac{16}{21}+1\)
\(A=5+\frac{3}{4}+\frac{16}{21}\)
\(B=\left(27,5\times0,1+2,5\times0,1\right)\times2\)
\(B=\left[\left(27,5+2,5\right)\times0,1\right]\times2\)
\(B=\left(30\times0,1\right)\times2\)
\(B=5\)
\(\Rightarrow5+\frac{3}{4}+\frac{16}{21}>5\)
\(\Rightarrow A>B\)
1a
75/100+18/21+19/32+1/4+3/21+13/32
= 3/4 +6/7+19/32+1/4+1/7+13/32
= (3/4+1/4)+(19/32+13/32)+(6/7+1/7)
= 1+1+1=3
1b
22/5+51/9+11/4+3/5+1/3+1/4
=22/5+17/3+11/4+3/5+1/3+1/4
=(22/5+3/5)+(17/3+1/3)+(11/4+1/4)
=25/5+18/3+12/4
=5+6+3
=14
Ta có 12 + 22 + 32 + …102 = 385
Suy ra ( 12 +22 + 32 +…+102 ) .32 = 385.32
Do đó ta tính được A = 32 + 62 + 92 + …+302 = 3465
32.22-32.19
=33.(22-19)
=32.3
=33
=27