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TL :
Ko biết thì đừng làm
Nhớ làm hết , chi tiết mới đc 1 SP
HT
a) \(A=\frac{135}{135.136-1}\) và \(B=\frac{136}{136.137-1}\)
\(A=\frac{1}{136-1}=\frac{1}{135}\) \(B=\frac{1}{137-1}=\frac{1}{136}\)
Vì \(\frac{1}{136}\)< \(\frac{1}{135}\)nên A > B.
a, A = \(\frac{136-1}{\left(136-1\right)136-1}\) = \(\frac{136-1}{136^2-136-1}\) B=\(\frac{136}{136\left(136+1\right)-1}\)=\(\frac{136}{136^2+136-1}\)
x=136, A-B =\(\frac{x-1}{x^2-x-1}\)-\(\frac{x}{x^2+x-1}\) =\(\frac{x^3+x^2-x-x^2-x+1-x^3+x^2+x}{\left(x^2-1\right)^2-x^2}\)=\(\frac{x^2-x+2}{\left(x^2-1\right)^2-x^2}\)<0
=> A<B
b,A = \(\frac{456-333}{456}\)= 1-333/456 B=\(\frac{789-333}{789}\)= 1-333/789
=> A>B
c, 3/14<3/13<3/12<3/11<3/10 <2/5
M = 3/10+3/11+3/12+3/13+3/14 < 2/5 x5 = 2= N
Ta có:
\(3A=3+3^2+3^3+...+3^{11}\)
\(3A-A=2A=3^{11}-1\)
\(2A=\frac{3^{11}-1}{2}< B=\frac{3^{11}}{2}\)
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) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(N< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(N< 1-\frac{1}{100}\)
\(N< \frac{99}{100}< \frac{75}{100}=\frac{3}{4}\)
\(a,\)
Để A là phân số thì \(n-2\ne0\Rightarrow n\ne2\)
b, Ta có :
\(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)
Mà \(3⋮n+2\Rightarrow n+2\inƯ(3)=\left\{\pm1;\pm3\right\}\)
Tự xét bảng
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