So sánh :
A=\(1+3+3^2+3^3+...+3^{10}\) và B=\(\frac{3^{11}}{2}\)
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Đặt C = A - 3 = \(1+3+3^2+...+3^{10}\)
\(\Rightarrow3C=3+3^2+3^3+...+3^{11}\)
\(\Rightarrow2C=3C-C=3^{11}-1\)
\(\Rightarrow C=\frac{3^{11}-1}{2}\) \(\Rightarrow A-3=\frac{3^{11}-1}{2}=\frac{3^{11}}{2}-\frac{1}{2}\)
\(\Rightarrow A=\frac{3^{11}}{2}-\frac{1}{2}+3=\frac{3^{11}}{2}+\frac{5}{2}>\frac{3^{11}}{2}=B\)
Vậy A > B
A = 1 + 3 + 32 + 33 + ....... + 310 và B = 311 / 2
Ta có A = 1 + 3 + 32 +....+ 310
3A = 3. ( 1 + 3 + .... + 310 )
3A = 3 + 32 + 33 +.......+ 311
3A - A = (3 + 32 + 33+ ...+ 311)- ( 1 + 3 + ....+ 310)
2A = 311 - 1
A = 311 - 1 / 2 thì < 311 / 2
=> A < B
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
a)
Có: \(2>1>0\)
\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)
b) Có: \(0< \sqrt{3}< 3\)
\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)
c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)
\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)
d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)
\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)
a: 2=1+1<1+căn 2
b: 4=1+3>1+căn 3
c: -2căn 11=-căn 44
-10=-căn 100
mà 44<100
nên -2 căn 11>-10
d: 12=3*4=3*căn 16>3*căn 11
TL :
Ko biết thì đừng làm
Nhớ làm hết , chi tiết mới đc 1 SP
HT
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
So Sánh: B = \(\frac{^{3^{10}.11+3^{10}.5}}{3^9.2^4}\) và C= \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
Ta có:
B=\(\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) = \(\frac{3^{10}.16}{3^9.2^4}\)= \(\frac{3^9.3.16}{3^9.16}\)= 3
C=\(\frac{2^{10}.\left(13+65\right)}{2^8.104}\) =\(\frac{2^{10}.78}{2^8.104}\) = \(\frac{2^8.2^2.78}{2^8.104}\)= \(\frac{4.78}{104}\) = \(\frac{4.78}{4.26}\)=\(\frac{78}{26}\)=3
=>B=C
Ta có:
\(3A=3+3^2+3^3+...+3^{11}\)
\(3A-A=2A=3^{11}-1\)
\(2A=\frac{3^{11}-1}{2}< B=\frac{3^{11}}{2}\)