\(\dfrac{x}{6}-\dfrac{1}{y}=\dfrac{1}{2}\) a.
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\) b.
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30 tháng 6 2021
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\)
\(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{25}{12}\)
\(\dfrac{1}{2}:y=\dfrac{25}{12}:\dfrac{3}{5}\)
\(\dfrac{1}{2}:y=\dfrac{125}{36}\)
\(y=\dfrac{1}{2}:\dfrac{125}{36}\)
\(y=\dfrac{18}{125}\)
b) \(\dfrac{4}{3}-\dfrac{1}{2}\times y=1\)
\(\dfrac{1}{2}\times y=\dfrac{4}{3}-1\)
\(\dfrac{1}{2}\times y=\dfrac{1}{3}\)
\(y=\dfrac{1}{3}:\dfrac{1}{2}\)
\(y=\dfrac{2}{3}\)
c) \(\dfrac{1}{4}+y:\dfrac{1}{3}=\dfrac{5}{6}\)
\(y:\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{4}\)
\(y:\dfrac{1}{3}=\dfrac{7}{12}\)
\(y=\dfrac{7}{12}\cdot\dfrac{1}{3}\)
\(y=\dfrac{7}{36}\)
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
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
Bài 4:
\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{5}{8}\)
________________
\(456+y:87=23987\)
\(\Rightarrow y:87=23987-456\)
\(\Rightarrow y:87=23531\)
\(\Rightarrow y=23531\cdot87\)
\(\Rightarrow y=2047197\)
18 tháng 7 2023
a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)
\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)
\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)
b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\dfrac{17}{15}\)
\(=\dfrac{59}{30}\)
Bài 2:
a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)
\(y=\dfrac{5}{8}\)
b)\(456+y:87=23987\)
\(y:87=23987-456\)
\(y:87=23531\)
\(y=23531\times87\)
\(y=2047197\)
MH
9 tháng 5 2022
Bài 1:
+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)
\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)
+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)
\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)
\(y=\dfrac{9}{50}\)
HQ
Hà Quang Minh
Giáo viên
10 tháng 1
\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)
\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)
\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)
\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)
8 tháng 1 2018
\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)
Thay vào (1) ta được :
\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}
\(b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\) (2)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = 1/x ; b = 1/y
Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{4};-3\) )}
c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)
ĐK xác định : x≠0 ; y ≠0
Áp dụng quy tác cộng đại số ta có :
\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)
ĐK xác định : x≠0 ; y≠0
áp dụng quy tắc cộng đại số ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)
Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}
e) ĐK xác định x≠0 ; y≠0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)
Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
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 = \(\dfrac{15}{13}\)
