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đặt x/3=y/=k(k khác 0) =>x=3k;y=7k
=>x.y=3k.7k=21.k^2=84
=>k^2=4=(2)^2 hoặc(-2)^2
th1:k=2=> x=6;y=14
th2:k=-2 =>x=-6;y=-14
Đặt \(\frac{x}{3}=\frac{y}{7}=k\) ta có :
\(x=3k\) ;\(y=7k\)
Vì \(x.y=84\Rightarrow3k.7k=21k^2=84\)
\(\Rightarrow k^2=4=2^2\)
\(\Rightarrow\orbr{\begin{cases}k=-2\\k=2\end{cases}}\)
+TH1: \(k=-2\Rightarrow\hept{\begin{cases}x=-6\\y=-14\end{cases}}\)
+TH2: \(k=2\Rightarrow\hept{\begin{cases}x=6\\y=14\end{cases}}\)
Vậy (x,y) = {(-6,-14);(6,14)}
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\(\frac{x}{3}=\frac{y}{7}\) va xy=84
Dat : \(\frac{x}{3}=\frac{y}{7}=k\)
x.y=21k2
84 =21k2
k2 = 4
k = +-2
Neu : k=4\(\Rightarrow x=4.3=12;y=4.7=28\)
Neu : k=-4\(\Rightarrow x=-4.3=-12;y=-4.7=-28\)
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Ta có : \(\frac{x}{3}=\frac{y}{7}\) và x.y = 84
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x.y}{3.7}=\frac{84}{21}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{3}=4\Rightarrow x=4.3=12\\\frac{y}{7}=4\Rightarrow y=4.7=28\end{matrix}\right.\)
Vậy....
\(\dfrac{x}{3}=\dfrac{y}{7}\) và x.y=84
Ta có:\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x.y}{3.7}=\dfrac{84}{21}=4\)
\(\Rightarrow\dfrac{x}{3}=4\Rightarrow x=3.4=12\)
\(\dfrac{y}{7}=4\Rightarrow y=7.4=21\)
Lần sau có dạng giống vậy thì bạn áp dụng vào để tính nhé!
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\(\frac{x}{5}=\frac{y}{7}=k\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)
\(x\cdot y=140\)
\(\Rightarrow5k\cdot7k=140\)
\(\Rightarrow35k^2=140\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(k=2\Rightarrow\hept{\begin{cases}x=2\cdot5=10\\y=2\cdot7=14\end{cases}}\)
\(k=-2\Rightarrow\hept{\begin{cases}x=-2\cdot5=-10\\y=-2\cdot7=-14\end{cases}}\)
\(7x=3y\)
\(\Rightarrow\frac{x}{3}=\frac{y}{7}=k\Rightarrow\hept{\begin{cases}x=3k\\y=7k\end{cases}}\)
\(\Rightarrow x\cdot y=3k\cdot7k=2100\)
\(\Rightarrow21k^2=2100\)
\(\Rightarrow k^2=100\)
\(\Rightarrow k=\pm10\)
\(k=10\Rightarrow\hept{\begin{cases}x=10\cdot3=30\\y=10\cdot7=70\end{cases}}\)
\(k=-10\Rightarrow\hept{\begin{cases}x=-10\cdot3=-30\\y=-10\cdot7=-70\end{cases}}\)
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ta có \(\frac{x\left(x.y\right)}{y\left(x.y\right)}=\frac{3}{10}:\left(-\frac{3}{50}\right)=-5=\frac{x}{y}\)
\(x=-5y\)suy ra \(-5\left(-5y-y\right)=\frac{3}{10}\)suy ra \(30y^2=\frac{3}{10}\)
nên \(y=\frac{1}{10}\)hoặc \(y=-\frac{1}{10}\)
+) Với \(y=\frac{1}{10}\)suy ra \(x=-5.\frac{1}{10}=-\frac{1}{2}\)
+) Với \(y=-\frac{1}{10}\)suy ra \(x=-5.\left(-\frac{1}{10}\right)=\frac{1}{2}\).
Chúc làm bài may mắn
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Có: \(\frac{3}{x}=\frac{y}{7}=\frac{x}{3}=\frac{y}{7}\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Thay \(x=3k;y=7k\) vào \(x.y=84\), ta có:
\(3k.7k=84\\ \Leftrightarrow21k^2=84\\ \Leftrightarrow k^2=4\\ \Leftrightarrow k^2=\left(\pm2\right)^2\\ \Rightarrow k\in\left\{2;-2\right\}\)
+Khi \(k=2\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
+Khi \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-2.3=-6\\y=-2.7=-14\end{matrix}\right.\)
Vậy...
Ta có: \(\frac{3}{x}=\frac{y}{7}.\)
\(\Rightarrow\frac{x}{3}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
Có: \(x.y=84\)
=> \(3k.7k=84\)
=> \(21k^2=84\)
=> \(k^2=84:21\)
=> \(k^2=4\)
=> \(k=\pm2.\)
TH1: \(k=2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.7=14\end{matrix}\right.\)
TH2: \(k=-2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right).3=-6\\y=\left(-2\right).7=-14\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;14\right),\left(-6;-14\right).\)
Chúc bạn học tốt!
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a Ta có: \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\left(1\right)\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\left(2\right)\)
Từ (1);(2) => \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
=> x = 2 x 10 = 20
y = 2 x 15 = 30
z = 2 x 21 = 42
b) Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
=> x = 2k ; y = 3k
=> xy = 6.k2
=> 54 = 6.k2
=> k2 = 54 : 6 = 9
=> k = 3 hoặc k = -3
=> x = 3 x 2=6 hoặc x =( -3) x 2 = -6
y = 3 x 3 = 9 hoặc y = (-3) x 3 = -9
\(\text{a,Ta có:}\)\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\) \(\text{và}\)\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
\(\text{Áp dụng tính chất DTSBN có}\)
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\text{Suy ra}:x=2.10=20;y=2.15=30;z=2.21=42\)
\(\text{Vậy }x=20;y=30;z=42\)
\(\text{b, Đặt }\frac{x}{2}=\frac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(\text{Theo đề, ta có}\)
\(xy=54\Rightarrow2k.3k=54\Rightarrow6k^2=54\Rightarrow k^2=9\Rightarrow k=3\text{hoặc }k=-3\)
\(\text{Suy ra: }x=2.3=6\text{hoặc}x=2.\left(-3\right)=-6\) \(y=3.3=9\text{ hoặc }y=-3.3=-9\)
\(\text{Vậy với k=3 }\Rightarrow x=6;y=9\)
\(\text{với k=-3\Rightarrow x=-6;y=-9}\)
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Đặt: \(\frac{x}{4}=\frac{y}{7}=k\)
\(\Rightarrow x=4k;y=7k\)
\(\Rightarrow xy=4k.7k=112\)
\(\Rightarrow28k^2=112\)
\(\Rightarrow k^2=\frac{112}{28}=4\)
\(\Rightarrow\left[\begin{array}{nghiempt}k=2\\k=-2\end{array}\right.\)
Với \(k=2\) \(\Rightarrow\left[\begin{array}{nghiempt}x=4k=2.4=8\\y=7k=2.7=14\end{array}\right.\)
Với \(k=-2\Rightarrow\left[\begin{array}{nghiempt}x=4k=-2.4=-8\\y=7k=-2.7=-14\end{array}\right.\)
Đặt \(\frac{x}{4}=\frac{y}{7}\) = k
=> x = 4k; y = 7k
Ta thay vào: x . y = 112
=> 4k . 7k = 112
=> 28 . k2 = 112
=> k2 = 112 : 28
=> k2 = 4
=> k = 2 hoặc k = -2
Nếu k = 2 => x = 4 . 2 = 8; y = 7k = 7 . 2 = 14
Nếu k = -2 => x = 4 . (-2) = -8; y = 7 . (-2) = -14
Vậy x = {-8; 8} và y = {-14; 14}
Ta có: xy = 84
=> \(y=\frac{84}{x}\)
=> \(\frac{x}{3}=\frac{\frac{84}{x}}{7}\)
=> \(\frac{x}{3}=\frac{12}{x}\)
=> \(x^2=3.12=36\)
=> \(x=\pm6\)
Khi x = 6
=> \(y=\frac{84}{x}=\frac{84}{6}=14\)
Khi x = -6
=> \(y=\frac{84}{x}=\frac{84}{-6}=-14\)
Theo bài ra ta có: \(\frac{x}{3}=\frac{y}{7}\Rightarrow\frac{x}{3}.\frac{x}{3}=\frac{y}{7}.\frac{y}{7}=\frac{x}{3}.\frac{y}{7}\)
\(\Rightarrow\frac{x^2}{9}=\frac{y^2}{49}=\frac{84}{21}=4\)
\(\Rightarrow x^2=4.9=36\Rightarrow x=\pm6\)
\(\Rightarrow y^2=196=\pm14\)
Vậy \(x=\pm6\)
\(y=\pm14\)