1 phần bốn trừ 1 phần mười hai
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a; \(\dfrac{3}{5}\) - \(\dfrac{-7}{10}\) - \(\dfrac{13}{-20}\)
= \(\dfrac{12}{20}\) + \(\dfrac{14}{20}\) + \(\dfrac{13}{20}\)
= \(\dfrac{39}{20}\)
b; \(\dfrac{1}{2}\) + \(\dfrac{1}{-3}\) + \(\dfrac{1}{4}\) - \(\dfrac{-1}{6}\)
= \(\dfrac{6}{12}\) - \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{2}{12}\)
= \(\dfrac{7}{12}\)
\(\frac{4}{7}-\frac{2}{15}\div\frac{4}{25}\)
\(=\frac{4}{7}-\frac{2}{15}\times\frac{25}{4}\)
\(=\frac{4}{7}-\frac{50}{60}\)
\(=\frac{4}{7}-\frac{5}{6}\)
\(=\frac{24}{42}-\frac{35}{42}\)
\(=-\frac{11}{42}\)
Bài 1:
(\(x-12\))80 + (y + 15)40 = 0
Vì (\(x-12\))80 ≥ 0 ∀ \(x\); (y + 15)40 ≥ 0 ∀ y
Vậy (\(x-12\))80 + (y + 15)40 = 0
⇔ \(\left\{{}\begin{matrix}x-12=0\\y+15=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=12\\y=-15\end{matrix}\right.\)
Vậy \(\left(x;y\right)\) = (12; -15)
Bài 2:
\(\dfrac{x}{y}\) = \(\dfrac{a}{b}\) (đk \(y;b\ne0\))
⇒ \(\dfrac{x}{a}\) = \(\dfrac{y}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a}\) = \(\dfrac{y}{b}\) = \(\dfrac{x-y}{a-b}\)
⇒ \(\dfrac{x}{a}\) = \(\dfrac{x-y}{a-b}\)
⇒ \(\dfrac{x-y}{x}\) = \(\dfrac{a-b}{a}\) (đpcm)
a; 0,2.\(\dfrac{15}{36}\) - (\(\dfrac{2}{5}\) + \(\dfrac{2}{3}\)): 1%
= \(\dfrac{1}{12}\) - \(\dfrac{16}{15}\): \(\dfrac{1}{100}\)
= \(\dfrac{1}{12}\) - \(\dfrac{320}{3}\)
= \(\dfrac{1}{12}\) - \(\dfrac{1280}{12}\)
= - \(\dfrac{1279}{12}\)
b; 75% - 1\(\dfrac{1}{2}\) + 0,5 : \(\dfrac{5}{12}\)
= 0,75 - 1,5 + 1,2
= -0,75 + 1,2
= 0,45
c; 1\(\dfrac{3}{15}.0,75-\left(\dfrac{8}{15}+0,25\right)\).\(\dfrac{24}{47}\)
= \(\dfrac{28}{15}\).0,75 - \(\dfrac{47}{60}\).\(\dfrac{24}{47}\)
= \(\dfrac{7}{5}-\dfrac{2}{5}\)
= 1
d; \(\dfrac{32}{15}\): (-1\(\dfrac{1}{5}\) + 1\(\dfrac{1}{3}\))
= \(\dfrac{32}{15}\): (-\(\dfrac{6}{5}\) + \(\dfrac{4}{3}\))
= \(\dfrac{32}{15}\): \(\dfrac{2}{15}\)
= 16
1+ \(\frac{10}{4}\)+ \(\frac{10}{12}\)+ \(\frac{10}{24}\)+ \(\frac{10}{40}\)+ \(\frac{10}{60}\)= \(\frac{31}{6}\).
1/4 - 1/12 = 3/12 - 1/12 = 2/12 = 1/6