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\(\dfrac{-18}{24}+\dfrac{15}{-21}=\dfrac{-3}{4}+\dfrac{-5}{7}=\dfrac{-21}{28}+\dfrac{-20}{28}=\dfrac{-41}{28}\)
\(\dfrac{-18}{24}+\dfrac{15}{-21}=\dfrac{-3}{4}+\dfrac{-5}{7}=\dfrac{-21}{28}+\dfrac{-20}{28}=\dfrac{-21+\left(-20\right)}{28}=\dfrac{-41}{28}\)
2+(4-6-8+10) +(12-14-16+18) +.......-2008
2+0+0+....-2008
2+2008=2010
Phần 1);2) dễ nha nên bạn tự thay vào tính nha nên mk làm phần 3)
C=(a.a+2.a.b+b.b)
C=(a2+2ab +b2)
C=(a.(a+2b)+b2)
Chúc bn học tốt
\(A=\left(\frac{2.18-1}{18}\right)\left(\frac{2.18-2}{18}\right)\left(\frac{2.18-3}{18}\right)....\left(\frac{2.18-35}{18}\right)\left(\frac{2.18-36}{18}\right)\left(\frac{2.18-37}{18}\right)...\left(\frac{2.18-100}{18}\right)\)
\(=\frac{35}{18}.\frac{34}{18}.\frac{33}{18}...\frac{1}{18}.\frac{0}{18}.\frac{-1}{18}...\frac{-64}{18}=0\)
3^3 . 18 + 72 . 4^2 - 41 . 18
= 27.18 + 72.16 - 41.18
= 27.18 + 18.4.16 - 41.18
=18.(27+4.16-41)
=18.(27+64-41)
=18.50
=900
\(\Leftrightarrow\dfrac{x}{-2}=\dfrac{1}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}\)
=>x=1; y=36; z=12
a; (15 - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) : \(\dfrac{51}{40}\)
(\(\dfrac{270}{18}\) - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) x \(\dfrac{40}{51}\)
= \(\dfrac{149}{18}\) : \(\dfrac{297}{27}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{18}\) x \(\dfrac{27}{297}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{198}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{198}\) - \(\dfrac{330}{198}\)
= \(\dfrac{-181}{198}\)
b; (- 3,2) x (- \(\dfrac{15}{64}\)) + (0,8 - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= (\(\dfrac{-16}{5}\)) x ( \(\dfrac{-15}{64}\)) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + (\(\dfrac{12}{15}\) - \(\dfrac{34}{15}\)) : \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + \(\dfrac{-22}{15}\) : \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) - \(\dfrac{22}{15}\) x \(\dfrac{3}{11}\)
= \(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)
= \(\dfrac{15}{20}\) - \(\dfrac{8}{20}\)
= \(\dfrac{7}{20}\)
\(2x^2\cdot3^y=18\)
\(\Leftrightarrow2x^2\cdot3^y=2\cdot3^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{1;-1\right\}\\y=2\end{matrix}\right.\)
1 - 1 cũng không biết là thua luôn
1-1=0