-60 :4.3 -(-3)^3 :(-1)^2024 -|-15|
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a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
Ta có:
Mẫu số chung 2 phân số: 84
\(\dfrac{3}{7}=\dfrac{3*12}{7*12}=\dfrac{36}{84}\)
\(\dfrac{5}{12}=\dfrac{5*7}{12*7}=\dfrac{35}{84}\)
Vì \(36>35\) nên\(\dfrac{36}{84}>\dfrac{35}{84}\)
Vậy \(\dfrac{3}{7}>\dfrac{5}{12}\)
Ta có:
\(\dfrac{9}{8}>1>\dfrac{2023}{2024}\) nên \(\dfrac{9}{8}>\dfrac{2023}{2024}\)
Ta có:
\(\dfrac{1+15}{16}=1\)
\(\dfrac{1+16}{15}=\dfrac{17}{15}>1\)
\(\Rightarrow\dfrac{1+15}{16}>\dfrac{1+16}{15}\)
Lời giải:
a.
$=5^{6-3}+2^{3-2}=5^3+2^1=125+2=127$
b.
$=3333:3+225:15=1111+15=1126$
c.
$=15.2^3+4.3^2-5.7=15.8+4.9-35=120+36-35=120+1=121$
d.
$=5.4^2-18:3^2=5.16-18:9=80-2=78$
e.
$=6^2:4.3+2.5^3=36:4.3+2.125$
$=9.3+250=27+250=277$
f.
$=80-(4.25-3.8)=80-(100-24)=80-100+24=-20+24=4$
\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)
Ta có :
\(\dfrac{1}{k+1}C^{2k-1}_n=\dfrac{1}{k+1}.\dfrac{n!}{\left(2k-1\right)!\left(n-2k+1\right)!}\)
\(=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{2k!\left[\left(n+1\right)-2k\right]!}\)
\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)
\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)
\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)
-60 :4.3-(-3)^3:(-1)^2014-|-15|
=-60:4.3+27:(-1)-15
=45+(-27)-15
=3