\(\frac{y}{13}\)+ \(\frac{6}{y}\)= \(\frac{1}{2}\)
b) \(\frac{24}{y}\):\(\frac{8}{3}\)=\(\frac{23}{20}\)
Tìm y
Những câu hỏi liên quan
1Tính.frac{18,4.0,3.7,5.4,8}{2,5.4,6.9,6.0,9}2.Tĩm x,y,zfrac{x}{13}frac{-15}{39}frac{20}{y}b.frac{-7}{x}frac{14}{16}frac{y}{32}c.frac{24}{x}frac{y}{32}frac{-6}{z}frac{3}{2}d.frac{-10}{15}frac{x}{-9}frac{-8}{y}frac{z}{-21}3.Tìm x, biết:frac{x+1}{6}frac{8}{3}Mong các bạn giúp mình!
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1Tính
.\(\frac{18,4.0,3.7,5.4,8}{2,5.4,6.9,6.0,9}\)
2.Tĩm x,y,z
\(\frac{x}{13}=\frac{-15}{39}=\frac{20}{y}\)
b.\(\frac{-7}{x}=\frac{14}{16}=\frac{y}{32}\)
c.\(\frac{24}{x}=\frac{y}{32}=\frac{-6}{z}=\frac{3}{2}\)
d.\(\frac{-10}{15}=\frac{x}{-9}=\frac{-8}{y}=\frac{z}{-21}\)
3.Tìm x, biết:
\(\frac{x+1}{6}=\frac{8}{3}\)
Mong các bạn giúp mình!
\(\text{3 Giải}\)
\(\frac{x+1}{6}=\frac{8}{3}=\frac{16}{6}\Rightarrow x+1=16\Rightarrow x=15.\text{Vậy: x=15}\)
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Cho ba số x,y,z sao cho \(\frac{x}{3} = \frac{y}{4};\frac{y}{5} = \frac{z}{6}\)
a) Chứng minh: \(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}}\)
b) Tìm ba số x,y,z biết x – y + z = - 76
a) Ta có:
\(\begin{array}{l}\frac{x}{3} = \frac{y}{4} \Rightarrow \frac{x}{3}.\frac{1}{5} = \frac{y}{4}.\frac{1}{5} \Rightarrow \frac{x}{{15}} = \frac{y}{{20}};\\\frac{y}{5} = \frac{z}{6} \Rightarrow \frac{y}{5}.\frac{1}{4} = \frac{z}{6}.\frac{1}{4} \Rightarrow \frac{y}{{20}} = \frac{z}{{24}}\end{array}\)
Vậy \(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}}\) (đpcm)
b) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}} = \frac{{x - y + z}}{{15 - 20 + 24}} = \frac{{ - 76}}{{19}} = - 4\)
Vậy x = 15 . (-4) = -60; y = 20. (-4) = -80; z = 24 . (-4) = -96
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\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\) (x=0,5;y=3)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)
\(B=\frac{x}{y}-\frac{3}{2}\)
Thế x = 0, 5 = 1/2 ; y = 3 ta được :
\(B=\frac{\frac{1}{2}}{3}-\frac{3}{2}=\frac{1}{6}-\frac{9}{6}=-\frac{8}{6}=-\frac{4}{3}\)
Ta có:\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)(Do\(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\ne0\))
\(B=\frac{x}{y}-\frac{3}{2}\)
Thay x = 0,5; y = 3 vào B ta được:
\(B=\frac{0,5}{3}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{9}{6}\)
\(B=-\frac{4}{3}\)
Vậy\(B=-\frac{4}{3}\)tại x = 0,5; y = 3
Linz
Tìm x, y, z biết:
a, frac{x}{3}frac{y}{4}frac{z}{5}và x+y-24
b, frac{x}{7}frac{y}{6}frac{z}{5}và 3z-2y20
c, frac{x}{2}frac{y}{3}frac{z}{4}và x+2y-3z-20
d, frac{x}{2}frac{y}{3};frac{y}{8}frac{z}{10}và x+y-z20
e, 3x2y;frac{y}{6}frac{z}{7}và x+y-z30
f, frac{x}{2}frac{y}{3}và xy 5400
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Tìm x, y, z biết:
a, \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}v\)à x+y=-24
b, \(\frac{x}{7}=\frac{y}{6}=\frac{z}{5}\)và 3z-2y=20
c, \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và x+2y-3z=-20
d, \(\frac{x}{2}=\frac{y}{3};\frac{y}{8}=\frac{z}{10}\)và x+y-z=20
e, 3x=2y;\(\frac{y}{6}=\frac{z}{7}\)và x+y-z=30
f, \(\frac{x}{2}=\frac{y}{3}\)và xy= 5400
Mấy bài còn lại tương tự nhé cậu
1) \(\frac{24}{-12}=\frac{x}{5}=\frac{-y}{3}\)Tìm x và y
2) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-5}{25}\le\frac{x}{10}< \frac{-3}{4}+\frac{4}{14}+\frac{-2}{8}+\frac{-3}{5}+\frac{5}{7}\)Tìm x
3) \(\frac{8.x+18}{2.x+6}\)Tìm x
hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
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hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Tìm x ,y thỏa mãn:
\(|\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x|+|1,5-\frac{11}{17}+\frac{23}{13}y|=0\)
Lời giải :
Do \(VT\ge0\forall x;y\)nên ta có hệ :
\(\hept{\begin{cases}\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x=0\\1,5-\frac{11}{17}+\frac{23}{13}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{9}\\y=\frac{-377}{782}\end{cases}}\)
Vậy...
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Bài 1:Tìm các số nguyên x,y biết:
a) \(\frac{6}{2x+1}=\frac{2}{7}\)
b) \(\frac{24}{7x-3}=\frac{-4}{25}\)
c) \(\frac{4}{x-6}=\frac{y}{24}=\frac{-12}{18}\)
d) \(\frac{-1}{5}\le\frac{x}{8}\le\frac{1}{4}\)
e) \(\frac{x+46}{20}=x\frac{2}{5}\)
f) \(y\frac{5}{y}=\frac{86}{y}\)
a) 6/2X +1 = 2/7
=6 x 7 = 2(2x+1)
42=4X + 2
42 - 2 = 4X
40 = 4X
10 =X
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tìm x,y \(\in\)N ,biết :
a, \(\frac{3}{4}< \frac{x}{30}< \frac{y}{60}< \frac{4}{5}\)
b, \(\frac{-1}{2}< \frac{x}{24}< \frac{y}{12}< \frac{-3}{8}\)
c, \(\frac{-7}{8}< \frac{x}{38}< \frac{y}{72}< \frac{-5}{6}\)