tim cap so x,y biet:
\(\dfrac{1+3.y}{12}\)\(=\dfrac{1+5.y}{5.x}\)\(=\dfrac{1+7.y}{4.x}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 9 2023
\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)
\(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)
y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)
y = \(\dfrac{245}{64}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 9 2023
2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)
\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)
\(\dfrac{12}{5}\): y = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)
\(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)
y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)
y = \(\dfrac{15}{13}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
LN
25 tháng 8 2023
mình ko chép đề bài nha
a) \(\dfrac{16}{5}\): \(\dfrac{7}{3}\) : y =\(\dfrac{12}{7}\)
\(\dfrac{48}{35}\): y = \(\dfrac{12}{7}\)
y = \(\dfrac{48}{35}\): \(\dfrac{12}{7}\)
y = \(\dfrac{4}{5}\)
28 tháng 9 2021
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
4 tháng 1 2022
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
ND
Nguyễn Đức Trí
VIP
25 tháng 7 2023
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
13 tháng 3 2021
\(C=5x^3y^2-4x^3y^2+3x^2y^3+\dfrac{1}{2}x^2y^3+\dfrac{1}{3}x^4y^5-3x^4y^5-\dfrac{1}{7}\)
\(=x^3y^2+\dfrac{7}{2}x^2y^3-\dfrac{8}{3}x^4y^5-\dfrac{1}{7}\)