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\(\dfrac{3}{17}\) .\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\) + 2022\(\dfrac{3}{17}\)
= (\(\dfrac{3}{17}\).\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)
= \(\dfrac{3}{17}\).(\(\dfrac{6}{29}\) - \(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)
= \(\dfrac{3}{17}\).(-1) + 2022 + \(\dfrac{3}{17}\)
= (- \(\dfrac{3}{17}\) + \(\dfrac{3}{17}\)) + 2022
= 0 + 2022
= 2022
b; \(\dfrac{5}{6}\) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{2}\)).20%
= \(\dfrac{5}{6}\) - \(\dfrac{5}{6}\).\(\dfrac{1}{5}\)
= \(\dfrac{5}{6}-\dfrac{1}{6}\)
= \(\dfrac{4}{6}\)
= \(\dfrac{2}{3}\)
\(a.3^{500}=\left(3^5\right)^{100}=125^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}\)
\(V\text{ì}\)\(125^{100}< 343^{100}=>3^{500}< 7^{300}\)
\(99^{20}=\left(9^2\right)^{10}=81^{10}\)
Vì 8110 < 999910 => 9920 < 999910
bai2
UCLN (n,n+2)=d
=>(n+2)-n chia hết cho d
2 chia het cho d
vay d thuoc uoc cua 2={1,2}
nếu n chia hết cho 2 uoc chung lon nhta (n,n+2) la 2
neu n ko chia het cho 2=> (n,n+2) nguyen to cung nhau
BCNN =n.(n+2) neu n le
BCNN=n.(n+2)/2
Bài 3:
a: =>x/56=3/7
hay x=21
b: =>3/x=6/14
hay x=7
c: =>35/x=7/15
hay x=75
d: =>2x=-10/9
hay x=-10/18=-5/9