5 và 1 phần mười trừ 4 và 3 phần mười=???
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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; \(\dfrac{1}{2}\) + \(\dfrac{1}{-3}\) + \(\dfrac{1}{4}\) - \(\dfrac{-1}{6}\)
= \(\dfrac{6}{12}\) - \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{2}{12}\)
= \(\dfrac{7}{12}\)
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
\(\frac{7}{13}.\frac{5}{9}+\frac{7}{19}.\frac{8}{13}-3.\frac{7}{19}=\)\(\frac{-1288}{2223}\)
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)
\(\frac{7}{8}-\frac{7}{16}-\frac{11}{32}\)
= \(\frac{28}{32}-\frac{14}{32}-\frac{11}{32}\)
= \(\frac{28-14-11}{32}\)
= \(\frac{3}{32}\)
Đề : 7/8 - 7/16 - 11/32 = ?
Chọn MSC là 32 . Ta có phép tính : 28/32 - 14/32 - 11/32 = 3/32
Tíc nha !
bằng 4/5
\(5\frac{1}{10}-4\frac{3}{10}\)
\(=\frac{51}{10}-\frac{43}{10}\)
\(=\frac{8}{10}=\frac{4}{5}\)