1*1*1*1*...*1/1*2*3*4*...*123
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1 - 1/2 + 1/3 - 1/4 + .. + 1/121 - 1/122 + 1/123
= ( 1 + 1/3 + ... + 1/121 + 1/123 ) - ( 1/2 + 1/4 + .. + 1/122 )
Đến bước này ta sẽ cùng cộng 2 vế với : 1/2 + 1/4 + .. + 1/122
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/121 + 1/122 + 1/123 ) - 2. ( 1/2 + 1/4 + .. + 1/122 )
= ( 1 + 1/2 + 1/3 + ...+ 1/122 +1 /123 ) - ( 1 + 1/2 + ...+ 1/61 )
= 1/62 + 1/63 + ..+1/122 + 1/123
Chúc học giỏi !!
a) \(\dfrac{4}{5}-\left(\dfrac{1}{5}+\dfrac{1}{3}\right)=\dfrac{4}{5}-\left(\dfrac{3}{15}+\dfrac{5}{15}\right)=\dfrac{4}{5}-\dfrac{8}{15}=\dfrac{12}{15}-\dfrac{8}{15}=\dfrac{4}{15}\)
b) \(\dfrac{12}{7}-\left(\dfrac{3}{14}-1\right)=\dfrac{12}{7}-\left(-\dfrac{11}{14}\right)=\dfrac{12}{7}+\dfrac{11}{14}=\dfrac{5}{2}\)
c) \(\dfrac{11}{2}-9\cdot\left(\dfrac{7}{5}-\dfrac{1}{2}\right)=\dfrac{11}{2}-9\cdot\dfrac{9}{10}=\dfrac{11}{2}-\dfrac{81}{10}=\dfrac{55}{10}-\dfrac{81}{10}=-\dfrac{13}{5}\)
d) \(3\times9\left(\dfrac{4}{7}+\dfrac{3}{5}\right)=27\times\dfrac{41}{35}=\dfrac{1107}{35}\)
e) \(\left(\dfrac{1}{3}-\dfrac{1}{5}\right):2=\left(\dfrac{5}{15}-\dfrac{3}{15}\right):2=\dfrac{2}{15}:2=\dfrac{2}{15}\times\dfrac{1}{2}=\dfrac{1}{15}\)
f) \(\dfrac{123}{34}:\dfrac{1}{17}-\dfrac{21}{34}:\dfrac{1}{17}=\left(\dfrac{123}{34}-\dfrac{21}{34}\right):\dfrac{1}{17}=3:\dfrac{1}{17}=3\times17=51\)
2/5 x 1/2 : 1/3 = 1/5 : 1/3
= 3/5
2/9 : 2/3 x 1/2 = 1/3 x 1/2
= 1/6
2/7 : 2/3 - 1/7 = 3/7 - 1/7
= 2/7
1/2 x 1/3 + 1/4 = 1/6 + 1/4
= 5/12
5535 : 123 = 45 ; 6560 : 234 = 28 ( dư 8 ) ; 32076 : 123 = 260 ( dư 96 )
Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)
\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)
Lời giải:
a)
\(A=75^2+50.75+25^2=75^2+2.25.75+25^2\)
\(=(75+25)^2=100^2=10000\)
b) \(123^2+23^2-46.123=123^2-2.23.123+23^2\)
\(=(123-23)^2=100^2=10000\)
c) \((3^4-1)(3^4+1)-9^4\)
\(=[(3^4)^2-1^2]-9^4=(9^4-1)-9^4=-1\)
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= 2952