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Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
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\(\left(3n\right)^{100}\\ =3^{100}.n^{100}\\ =\left(3^4\right)^{25}.n^{100}\\ =81^{25}.n^{100}⋮81\)
Vậy \(\left(3n\right)^{100}⋮81\)
Chúc em học tốt!
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Giải:
Có:
\(S=\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)+4^3\)
Ta nhân thấy rằng trong tích \(\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)\) có một thừa số bằng 0, đó là thừa số \(2018-2018\)
Mà trong một tích, nếu có một thừa số bằng 0 thì tích đó bằng 0
\(\Leftrightarrow\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)=0\)
\(\Leftrightarrow S=\left(2018-1\right)\left(2018-2\right)...\left(2018-2018\right)+4^3=0+4^3=4^3=64\)
Vậy \(S=64\)
Chúc bạn học tốt!
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S= (2018-1)(2018-2) .... (2018-2017) . 0 +43
=> S= 0 + 43 (Trong 1 tích có 1 thừa số bằng 0 thì tích đó bằng 0);
=>S= 4.4.4=64;
Vậy S=64
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Chứng minh:4 = 5
-->Ta có
-20 = -20
<=> 25 - 45 = 16 - 36
=> 5^2 - 2.5.9/ 2 = 4^2 - 2.4.9/2
Cộg cả 2 vế với (9/2)^2 để xuất hiện hằg đẳg thức :
5^2 - 2.5.9/2 + (9/2)^2 = 4^2 - 2.4.9/2 + (9/2)^2
<=> (5 - 9/2)^2 = (4 - 9/2 )^2
=> 5 - 9/2 = 4 - 9/2
=> 5 = 4
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a,
= \(\dfrac{13}{24}+\dfrac{-1}{16}\)
= \(\dfrac{23}{48}\)
b,
= \(\dfrac{7}{3}:\dfrac{7}{8}\)
= \(\dfrac{8}{3}-\dfrac{2}{5}\)
= \(\dfrac{34}{15}\)
c,
25%= \(\dfrac{1}{4}\)
= \(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}\left(0,5=\dfrac{1}{2}\right).\dfrac{12}{5}\)
= \(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{6}{5}\)
= \(\dfrac{-5}{4}+\dfrac{6}{5}\)
= \(\dfrac{-1}{20}\)
a) \(\dfrac{13}{24}+\dfrac{11}{24}:\dfrac{-22}{3}\)
=\(\dfrac{13}{24}+\dfrac{-1}{16}\)
=\(\dfrac{26}{48}+\dfrac{-3}{48}\)
=\(\dfrac{23}{48}\)
b) \(\left(\dfrac{5}{2}+\dfrac{-1}{6}\right):\dfrac{7}{8}-\dfrac{2}{5}\)
=\(\left(\dfrac{15}{6}+\dfrac{-1}{6}\right):\dfrac{7}{8}-\dfrac{2}{5}\)
=\(\dfrac{7}{3}:\dfrac{7}{8}-\dfrac{2}{5}\)
=\(\dfrac{8}{3}-\dfrac{2}{5}\)
=\(\dfrac{40}{15}-\dfrac{6}{15}\)
=\(\dfrac{34}{15}\)
c) 25% -\(1\dfrac{1}{2}\) +0,5.\(\dfrac{12}{5}\)
=\(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
= \(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{6}{5}\)
=\(\dfrac{-5}{4}+\dfrac{6}{5}\)
=\(\dfrac{-5}{12}\)