ba phần x trừ 1 bằng 9 phần 12
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\(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{2}\)
⇒ \(\dfrac{x}{12}-\dfrac{10}{12}=\dfrac{1}{12}\)
⇒ \(x=11\)
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)
3/4 - (x - 3) × 2 = 3 1/3 + 2
3/4 - (x - 3) × 2 = 10/3 + 2
3/4 - (x - 3) × 2 = 16/3
(x - 3) × 2 = 3/4 - 16/3
(x - 3) × 2 = -55/12 (lớp 5 chưa học số âm)
Em xem lại đề nhé
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
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