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Câu a )
S = 5 + 52 +..... + 52012
=> S \(⋮5\)
S = 5 + 52 +..... + 52012
S = ( 5 + 53 ) + ( 52 + 54 ) + ........ + ( 52010 + 52012 )
S = 5 ( 1 + 52 ) + 52 ( 1 + 52 ) + ......... + 52010 ( 1 + 52 )
S = 5 x 26 + 52 x 26 + ................ + 52010 x 26
S = 26 ( 5 + 52 + .... + 52010 )
=> S\(⋮26\)
=>\(S⋮13\)( do 26 = 13 x 2 )
Do ( 5 , 13 ) = 1
=> \(S⋮5x13\)
=> \(S⋮65\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
a) \(\frac{x+2}{x-2}=\frac{3}{2}\Rightarrow2\left(x+2\right)=3\left(x-2\right)\)
\(\Rightarrow2x+4=3x-6\)
\(\Rightarrow-x=-10\)
Vậy x = 10
b) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=9\)
\(\Rightarrow19x+50=126\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=4\)
Vậy x = 4
c) \(2x\left(3y-2\right)+\left(3y-2\right)=-55\)
\(\Rightarrow\left(3y-2\right)\left(2x+1\right)=-55=1.\left(-55\right)=\left(-1\right).55=5.\left(-11\right)=\left(-5\right).11\)
Lập bảng xét là ra
a) \(\frac{x+2}{x-2}=\frac{3}{2}\)
\(\Rightarrow2.\left(x+2\right)=3.\left(x-2\right)\)
\(\Rightarrow2x+4=3x-6\)
\(\Rightarrow x=10\)