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A = \(\frac{1}{3}-\frac{3}{4}-\frac{-3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}+\frac{-2}{9}\)
A = \(\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{73}\)
A = \(\left(\frac{3-2}{9}\right)-\left(\frac{27+1}{36}\right)+\left(\frac{9+1}{15}\right)+\frac{1}{73}\)
A = \(\frac{1}{9}-\frac{7}{9}+\frac{6}{9}+\frac{1}{73}\)
A = \(0+\frac{1}{73}=\frac{1}{73}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)
Ta có:\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)\(>B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\)
Mà \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}< 1\)
=>Đpcm
Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728
Làm
B = 1/3 - 3/4 - (-3)/5 + 1/73 - 1/36 + 1/15 + -2/9
B = 1/3 -3/4 + 3/5 +1/73 - 1/36 + 1/15 -2/9
B = [ 1/3 + 3/5 + 1/15 ] + [ -3/4 - 1/36 -2/9] + 1/73
B = [ 5/15 + 9/15 + 1/15 ] + [ -27/36 - 8/36 - 1/36 ] + 1/73
B = 1 + (-1) + 1/73
B = 1/73
HỌC TỐT Ạ
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x-101=0\)
\(\Leftrightarrow x=101\)
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}-\frac{100}{99}-\frac{99}{98}-\frac{96}{95}=0\)
\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)
\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)
\(\Leftrightarrow\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)
Do \(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\ne0\)
Mà \(x-101=0\Leftrightarrow x=101\)
Vậy x = 101
mình nhầm câu b:
Áp dụng....
A=10^11-1/10^12-1<10^11-1+11/10^12-1+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)
=10^10+1/10^11+1=B
Vậy A<B(câu này mới đúng còn câu b mình làm chung với câu a là sai)
a) Với a<b=>a+n/b+n >a/b
Với a>b=>a+n/b+n<a/b
Với a=b=>a+n/b+n=a/b
b) Áp dụng t/c a/b<1=>a/b<a+m/b+m(a,b,m thuộc z,b khác 0)ta có:
A=(10^11)-1/(10^12)-1=(10^11)-1+11/(10^12)-1+11=(10^11)+10/(10^12)+10=10.[(10^10)+1]/10.[(10^11)+1]
=(10^10)+1/(10^11)+1=B
Vậy A=B
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