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phần 1 = 24 x (31+42+27)=24 x 100=2400
phần 2 = (36 x (28+82)) + (64 x (69+41)) = (36x110) + (64x110) = 110 x (36+64) = 110x100=11000
\(2.31.12+4.6.41+8.27.3\)
\(=24.\left(31+41+27\right)\)
\(=24.99\)
\(=2376\)
\(2.31.12+4.6.41+8.27.3=24.31+24.41+24.27\)
\(=24.\left(31+41+27\right)=24.99=24.\left(100-1\right)\)
\(=24.100-24=2400-24=2376\)
2*31*12+4*6*41+8*27*3
= 24 *31 + 24 *41 + 24 * 27
= 24*( 31+41+27)
=24 * 99
=2376
2x31x12+4x6x42+8x27x3
=(2x31x12)+(4x6x42)+(8x27x3)
=(24x31)+(24x42)+(24x27)
=24x(31+42+27)
=24x100
=2400
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
\(\left(2^3.9^8.5\right):\left(9^2.\left(10-1\right)\right)\)
\(=2^3.9^8.5:9^3\)
\(=2^3.9^5.5\)
\(\left(2^3.9^4.9^3.45\right):\left(9^2.10-9^2\right)\)
(=) \(\left(2^3.3^8.3^6.5.3^2\right):9^2\left(10-1\right)\)
(=) \(\left(2^3.3^{15}.5\right):3^4.3^2\)
(=) \(2^3.3^9.5\)
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho
Bài 1 :
a) 40/49 > 15/21
b) 22/49 > 3/8
c) 25/46 < 12/18
a)Chịu
b)(68x8686 - 6868x86) xA = {[68x(86x101)] - [(68x101)x86]} xA =[(68x86x101) - (68x101x86)] xA = 0 xA = 0
a) 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= ( 2 x 12 ) x 31 + 24 x 42 + ( 8 x 3 ) x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 42 + 27 )
= 24 x 100
= 2400
b) ( 68 x 8686 - 6868 x 86) x ( 1 + 2 + 3 + ... + 2016 )
= ( 68 x 86 x 101 - 68 x 101 x 86 ) x ( 1 + 2 + 3 + ... + 2016 )
= 0 x ( 1 + 2 + 3 + ... +2016 )
= 0