so sanh:19/92*102+...+7/32*42+5/22*32+3/12*22 voi 11/10
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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
ta có: \(A=\frac{75}{100}+\frac{78}{21}+\frac{19}{32}+\frac{22}{21}+\frac{13}{32}\)
\(A=\frac{3}{4}+\left(\frac{78}{21}+\frac{22}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(A=\frac{3}{4}+4+\frac{16}{21}+1\)
\(A=5+\frac{3}{4}+\frac{16}{21}\)
\(B=\left(27,5\times0,1+2,5\times0,1\right)\times2\)
\(B=\left[\left(27,5+2,5\right)\times0,1\right]\times2\)
\(B=\left(30\times0,1\right)\times2\)
\(B=5\)
\(\Rightarrow5+\frac{3}{4}+\frac{16}{21}>5\)
\(\Rightarrow A>B\)
a) (-12).8 < (-19).3.
b) 11.(-2) > (-3).10.
c) (-16). 10 > (-32).11.
d) (-17).3 < (-22).2.
Lời giải:
\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)
\(=4A=4.385=1540\)
Ta có \(2^2+4^2+...+20^2=2^2\left(1^2+2^2+...+10^2\right)=2^2.385=1540\).
\(\frac{19}{9^2.10^2}+...+\frac{7}{3^2.4^2}+\frac{5}{2^2.3^2}+\frac{3}{1^2.2^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{8^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1< \frac{11}{10}\)