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a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)
hay \(\widehat{yOz}=40^0\)
Vậy: \(\widehat{yOz}=40^0\)
Mình nghĩ là lần sau bạn tách ra tầm khoảng 4-5 ý rồi đăng thì nó sẽ ngắn hơn ạ;-;. Chứ như này thì dài quá -> không có ai giúp đâu nhé:").
\(a,\dfrac{-6}{13}+\dfrac{7}{-13}=-\dfrac{13}{13}=-1\)
\(b,\dfrac{7}{12}+\dfrac{-9}{24}=\dfrac{14}{24}+\dfrac{-9}{24}=\dfrac{5}{24}\)
\(c,\dfrac{1}{5}+\dfrac{-5}{19}+\dfrac{4}{5}+\dfrac{-14}{19}=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{-5}{19}+\dfrac{-14}{19}\right)=1+\left(-1\right)=0\)
\(d,\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
\(=1+\left(-1\right)+\dfrac{-5}{7}=0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
\(e,\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}=\left(\dfrac{93}{11}+\dfrac{29}{8}\right)-\dfrac{38}{11}=\left(\dfrac{93}{11}-\dfrac{38}{11}\right)+\dfrac{29}{8}\)
\(=\dfrac{55}{11}+\dfrac{29}{8}=5+\dfrac{29}{8}=\dfrac{40}{8}+\dfrac{29}{8}=\dfrac{69}{8}\)
\(f,\dfrac{-1}{-5}+\dfrac{8}{7}+\dfrac{-6}{13}-1\dfrac{23}{7}-\dfrac{13}{24}=\dfrac{1}{5}+\dfrac{8}{7}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}\)
\(=\dfrac{7}{35}+\dfrac{40}{35}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{47}{35}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}\)
\(=\dfrac{611}{455}+\dfrac{-210}{455}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{401}{455}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{401}{455}-\dfrac{1950}{455}-\dfrac{13}{24}\)
\(=\dfrac{-1549}{455}-\dfrac{13}{24}=\dfrac{-37176}{10920}-\dfrac{5915}{10920}=\dfrac{-43091}{10920}=-3,946\) (số to quá bạn ơi;-;)
\(g,0,75+\dfrac{-1}{3}+\dfrac{3}{6^2}-\dfrac{5}{12}=\dfrac{3}{4}+\dfrac{-1}{3}+\dfrac{3}{36}-\dfrac{5}{12}=\dfrac{9}{12}+\dfrac{-4}{12}+\dfrac{1}{12}-\dfrac{5}{12}\)
\(=\dfrac{9+\left(-4\right)+1-5}{12}=\dfrac{1}{12}\)
\(h,\dfrac{-3}{7}\cdot\dfrac{-1}{9}+\dfrac{-7}{18}\cdot\dfrac{-3}{7}+\dfrac{5}{6}\cdot\dfrac{-3}{7}=\dfrac{-3}{7}\cdot\left(\dfrac{-1}{9}+\dfrac{-7}{18}+\dfrac{5}{6}+1\right)=\dfrac{-3}{7}\cdot\dfrac{4}{3}=\dfrac{-12}{21}\)
\(i,\dfrac{-5}{12}\cdot\dfrac{4}{19}+\dfrac{-7}{12}\cdot\dfrac{4}{19}-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(\dfrac{-5}{12}+\dfrac{-7}{12}\right)-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(-1\right)-\dfrac{40}{57}\)
\(=-\dfrac{4}{19}-\dfrac{40}{57}=-\dfrac{12}{57}-\dfrac{40}{57}=-\dfrac{52}{57}\)
\(j,\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+...+\dfrac{1}{99}\cdot\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{100}{200}-\dfrac{2}{200}=\dfrac{98}{200}=\dfrac{49}{100}\)
\(k,\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\)
\(=\dfrac{1}{2}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{6}+...+\dfrac{1}{98}\cdot\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{100}{200}-\dfrac{2}{200}=\dfrac{98}{100}=\dfrac{49}{100}\)
`@mt`
Hoc24.vn
*Do nhiều ý quá nên bài làm của tớ cũng không tránh khỏi nhầm lẫn, sai sót, có gì bạn tính lại nhé:(, tớ cảm ơn 😭.
a: =-6/13-7/13=-13/13=-1
b: =14/24-9/24=5/24
c: =1/5+4/5-5/19-14/19=1-1=0
d: \(=\dfrac{5}{13}+\dfrac{8}{13}-\dfrac{20}{41}-\dfrac{21}{41}-\dfrac{5}{7}=-\dfrac{5}{7}\)
e: \(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}=8+\dfrac{5}{8}=\dfrac{69}{8}\)
Xét ΔAND và ΔBMC có
AN=BM
ND=MC
AD=BC
=>ΔAND=ΔBMC
=>S vườn=S ABCD=90*70=6300m2
TL:
Số đó là 75
HT