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Đặt\(\frac{x}{3}=\frac{y}{7}=k\)
\(\Rightarrow\frac{x}{3}.\frac{y}{7}=k.k\Rightarrow\frac{xy}{21}=k^2\Rightarrow\frac{84}{21}=k^2\Rightarrow4=k^2\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)
Khi k = 2 thì: \(\frac{x}{3}=2\Rightarrow x=6;\frac{y}{7}=2\Rightarrow y=14\)
Khi k = -2 thì: \(\frac{x}{3}=-2\Rightarrow x=-6;\frac{y}{7}=-2\Rightarrow y=-14\)
Vậy: (x;y) = {(6; 14); (-6; -14)}
Đặt \(\frac{x}{3}=\frac{y}{7}=k\left(k>0\right)\)
=> x= 3k , y= 7k
Theo đề bài ta có : xy= 8 => 3k.7k= 84 => 21k2= 84 => k2= 4 => k= 2
=> x= 6, y= 14
\(x+y-xy+1=0\)
\(x+y-xy-1=-2\)
\(\Rightarrow x\left(1-y\right)-1\left(1-y\right)=-2\)
\(\Rightarrow\left(x-1\right)\left(1-y\right)=-2\)
\(\Rightarrow x-1;1-y\in U\left(-2\right)\)
\(U\left(-2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\Rightarrow x=2\\1-y=-2\Rightarrow y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\Rightarrow x=0\\1-y=2\Rightarrow y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=2\Rightarrow x=3\\1-y=-1\Rightarrow y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-2\Rightarrow x=-1\\1-y=1\Rightarrow y=0\end{matrix}\right.\end{matrix}\right.\)\
\(\dfrac{2}{x}-\dfrac{1}{9}=\dfrac{y}{3}\)
\(\Rightarrow\dfrac{2}{x}-\dfrac{1}{9}=\dfrac{3y}{9}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{3y}{9}+\dfrac{1}{9}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{3y+1}{9}\)
\(\Rightarrow x\left(3y+1\right)=18\)
\(\Rightarrow x;3y+1\in U\left(18\right)\)
Xét ước như bài trên
\(3x+3y-xy=0\)
\(\Rightarrow3x+3y-xy-9=-9\)
\(\Rightarrow x\left(3-y\right)-3\left(3-y\right)=-9\)
\(\Rightarrow\left(x-3\right)\left(3-y\right)=-9\)
\(\Rightarrow x-3;3-y\in U\left(9\right)\)
Xét ước ~~~
Bài 3:
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2y+z}{3\cdot3-2\cdot5+7}=\dfrac{84}{6}=14\)
=>x=42; y=70; z=98
\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)
\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)
\(\dfrac{1}{x}=\dfrac{1+2y}{6}\)
\(6=x\left(1+2y\right)\)
Tự làm típ
\(x\left(x+y\right)=\dfrac{1}{48};y\left(x+y\right)=\dfrac{1}{24}\)
\(x^2+xy=\dfrac{1}{48};xy+y^2=\dfrac{1}{24}\)
\(\Rightarrow x^2+xy-y^2-xy=\dfrac{1}{48}-\dfrac{1}{24}\)
\(x^2-y^2=\dfrac{-1}{24}\)
\(\left(x+y\right)\left(x-y\right)=\dfrac{-1}{24}\)(HĐT số 3)
Làm tips
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{2x}{8}=\dfrac{3y}{18}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{8}=\dfrac{3y}{18}=\dfrac{z}{5}=\dfrac{2x+3y-z}{8+18-5}=\dfrac{84}{21}=4\)
\(\Rightarrow\left[{}\begin{matrix}x=16\\y=24\\z=20\end{matrix}\right.\)
\(a,3x=-5y\Rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\) và \(y-x=-3\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{-5}=\dfrac{y}{3}=\dfrac{y-x}{3-\left(-5\right)}=-\dfrac{3}{8}\)
+) \(\dfrac{x}{-5}=\dfrac{3}{8}\Rightarrow8x=-15\Rightarrow x=-\dfrac{15}{8}\)
+) \(\dfrac{y}{3}=-\dfrac{3}{8}\Rightarrow8y=-9\Rightarrow y=-\dfrac{9}{8}\)
Vậy ...
xem lại đề
\(\)
Câu 1 :
a. Theo đề bài ta có :
\(\dfrac{x}{2}=\dfrac{y}{5}\) và \(x+y=21\)
Áp dụng t/c dãy tỉ số bằng nhau :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=3\Rightarrow x=2.3=6\\\dfrac{y}{5}=3\Rightarrow y=3.5=15\end{matrix}\right.\)
Vậy..............
b. Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}2k\\3y\end{matrix}\right.\)
mà \(x.y=54\)
hay \(2k.3k=54\)
\(\Rightarrow6.k^2=54\)
\(\Rightarrow k^2=9=\left(\pm3\right)^2\)
Với \(k=3\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=3.3=9\end{matrix}\right.\)
Với \(k=-3\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).3=-9\end{matrix}\right.\)
Vậy..............
c. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{12}{2}=6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=6\Rightarrow x=7.6=42\\\dfrac{y}{5}=6\Rightarrow y=5.6=40\end{matrix}\right.\)
Vậy............
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
Giải:
Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=k\) \(\Rightarrow\) \(\begin{cases}x=3k\\y=7k\end{cases}\)
Ta có:
\(xy=84\Rightarrow3k.7k=84\Rightarrow21k^2=84\)
\(\Rightarrow k^2=\dfrac{84}{21}=4\Leftrightarrow k=\) \(\pm 2\)
Ta có 2 trường hợp:
Trường hợp 1: Nếu \(k=2\) \(\Rightarrow\) \(\begin{cases}x=3.2=6\\y=7.2=14\end{cases}\)
Trường hợp 2: Nếu \(k=-2\) \(\Rightarrow\) \(\begin{cases}x=3.(-2)=-6\\y=7.(-2)=-14\end{cases}\)
Vậy...
Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{xy}{3y}=\dfrac{84}{3y}\)
=> \(\dfrac{y}{7}=\dfrac{84}{3y}\Rightarrow y\cdot3y=84\cdot7\Rightarrow3y^2=588\)
=> \(y^2=196\Rightarrow\left[{}\begin{matrix}y=14\\y=-14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{84}{14}=6\\x=\dfrac{84}{-14}=-6\end{matrix}\right.\)
Vậy.................