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\(5-\left(1997-2005\right)+1997\)
\(=5-1997+2005+1997\)
\(=2010\)
\(1579-\left(53+1579\right)-\left(-53\right)\)
\(=1579-53-1579+53\)
\(=0\)
\(-48-\left(357-48\right)+300\)
\(=\)\(-48-357+48+300\)
\(=-57\)
\(173-\left(36+173\right)+\left(175-27\right)-175-\left(-36+50\right)\)
\(=173-36-173+175-27-175+36-50\)
\(=-27-50\)
\(=-77\)
\(\)
\(-\left[-171+171+23\right]-\left[172-\left(38+172\right)\right]-49\)
\(=171-171-23-172+38+172-49\)
\(=-23+38-49=-34\)
a) Ta có: \(\frac{-697}{-313}=\frac{697}{313}>0\)
\(\frac{419}{-723}< 0\)
\(\Rightarrow\frac{419}{-713}< \frac{-697}{-313}\)
b) Ta có: \(\frac{190}{191}< 1\)
\(\frac{2019}{2018}>1\)
\(\Rightarrow\frac{190}{191}< \frac{2019}{2018}\)
c) Ta có: \(\frac{19}{27}< 1\)
\(\Rightarrow\frac{19}{27}< \frac{19.10+3}{27.10+3}\)
\(\Rightarrow\frac{19}{27}< \frac{193}{273}\)
d) Ta có: \(\frac{53}{47}< \frac{57}{47}\)
\(\frac{57}{43}>\frac{57}{47}\)
\(\Rightarrow\frac{53}{47}< \frac{57}{43}\)
a) 419/-723 < 0 < -697/-313
=> 419/-723 < -697/-313
b) 190/191 < 1 < 2019/2018
=> 190/191 < 2019/2018
d) 53/47 < 57/47 < 57/43
=> 53/47 < 57/43
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
mk làm phần b nhé
10n luôn có tổng các số hạng là 1
53 = 125
⇒ tổng các số hạng trong biểu thức là 126⋮9
⇒ 10n + 53 ⋮ 9 (đpcm)
a,Ta thấy :
\(\left\{{}\begin{matrix}60=2^2.3.5\\280=2^3.5.7\end{matrix}\right.\)
=> \(BCNN\left(60;280\right)=2^3.3.5.7=840\)
b,Ta thấy
\(\left\{{}\begin{matrix}84=2^2.3.7\\108=2^2.3^32^{ }\end{matrix}\right.\)
=> \(BCNN\left(84;108\right)=2^2.3^3.7=756\)
\(\)c, Ta thấy
\(\left\{{}\begin{matrix}13=13\\15=3.5\end{matrix}\right.\)
=> \(BCNN\left(13;15\right)=13.3.5=195\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)A=54-53/53+54=1/107=2/214
B=135-133/134+135=2/169
tự so sánh tiếp
a)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
\(A=\frac{5}{7}\cdot\frac{2}{3}\cdot\frac{7}{5}\cdot\frac{3}{2}\cdot\frac{19}{43}\)
\(=(\frac{5}{7}\cdot\frac{7}{5})\cdot(\frac{2}{3}\cdot\frac{3}{2})\cdot\frac{19}{43}\)
\(=1\cdot1\cdot\frac{19}{43}=\frac{19}{43}\)
Đáp án là D
Ta có: 171 + [(-53) + 96 + (-171)] = 171 + (-53) + 96 + (-171) = [171 + (-171)] + (-53) + 96
= 0 + (-53) + 96 = (-53) + 96 = 43