a) \(\dfrac{1}{2}x\dfrac{1}{3}+y=\dfrac{3}{4}\) b)\(\dfrac{1}{2}x\dfrac{1}{5}+y=\dfrac{3}{10}\)
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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
24 tháng 7 2023
1/2-2y=9/20
=>2y=1/2-9/20=1/20
=>y=1/20:2=1/40
b,3/5:4/3:y=2+7/10=9/20:y=27/10
=>y=9/20:27/10=1/6
c,y+y*3/2-y*1/2=1/10
=>y(1+3/2-1/2)=1/10
=>2y=1/10
=>y=1/10:2=1/20
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
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 7 2023
2: y \(\times\) \(\dfrac{3}{5}\) = \(\dfrac{9}{10}\)
2:y = \(\dfrac{9}{10}\) : \(\dfrac{3}{5}\)
2: y = \(\dfrac{3}{2}\)
y = 2 : \(\dfrac{3}{2}\)
y = \(\dfrac{4}{3}\)
\(\dfrac{5}{4}\) - \(\dfrac{2}{5}\) : y = 1
\(\dfrac{2}{5}\) : y = \(\dfrac{5}{4}\) - 1
\(\dfrac{2}{5}\): y = \(\dfrac{1}{4}\)
y = \(\dfrac{2}{5}\) : \(\dfrac{1}{4}\)
y = \(\dfrac{8}{5}\)
\(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{7}{2}\) - y) = \(\dfrac{3}{2}\)
\(\dfrac{7}{2}\) - y = \(\dfrac{3}{2}\) : \(\dfrac{3}{4}\)
\(\dfrac{7}{2}\) - y = 2
y = \(\dfrac{7}{2}\) - 2
y = \(\dfrac{3}{2}\)
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 = \(\dfrac{4}{5}\)
23 tháng 7 2021
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
9 tháng 11 2021
a) \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}=\dfrac{x\left(x+y\right)+2y^2-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2+xy+2y^2-x^2+xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{x-y}\)
b) \(B=\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}=\dfrac{x\left(x+2\right)-4x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c) \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3+1}=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)
a: \(y+\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)
=>\(y+\dfrac{1}{6}=\dfrac{3}{4}\)
=>\(y=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)
b: \(y+\dfrac{1}{2}\cdot\dfrac{1}{5}=\dfrac{3}{10}\)
=>\(y+\dfrac{1}{10}=\dfrac{3}{10}\)
=>\(y=\dfrac{3}{10}-\dfrac{1}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
tìm X ạ