Tìm x và y
\(\dfrac{x}{3}=\dfrac{y}{4}\)và\(x+y=14\)
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
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7 tháng 12 2021
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
7 tháng 12 2021
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
NM
0
LD
24 tháng 8 2018
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
HP
31 tháng 10 2021
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
11 tháng 11 2021
4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
18 tháng 10 2021
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
26 tháng 11 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)(nhận)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+2=1\\3+\dfrac{2}{x+y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x+y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=-1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)(nhận)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}y< >2x\\y< >-x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=6\\2x-y=3\end{matrix}\right.\)
=>x=2 và y=2x-3=4-3=1(nhận)
d:ĐKXĐ: \(\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)\)
e:
ĐKXĐ: \(x\ne\pm2y\)
\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{6}{x+2y}=5\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)\)
19 tháng 7 2021
a, Ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
\(\Rightarrow x=11;y=17;z=23\)
b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)
\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
\(\Rightarrow x=6;y=9;z=15\)
TG
19 tháng 7 2021
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
xyz = 810
=> 2k.3k.5k = 810
=> k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
VT
1 tháng 2 2023
a,Áp sụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)
1 tháng 2 2023
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\) (Vì 3x-2z=15)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)
Vậy ...
b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)
Vậy ...
c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\) (Vì 3z-2x=36)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)
Vậy ...
\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
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