tim x va y biet
1. x/3=y/4 va xy=192
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1, \(4x=5y\)
mà \(y-2x=-5\)
\(\Rightarrow x=\frac{y+5}{2}\)
\(\Rightarrow\left(\frac{y+5}{2}\right).4=5y\)
\(\Rightarrow\frac{4y+20}{2}=5y\)
\(\Rightarrow2y+10=5y\)
\(\Rightarrow10=3y\)
\(\Rightarrow y=\frac{10}{3}\)
\(\Rightarrow x=\frac{y+5}{2}=\frac{\frac{10}{3}+5}{2}=\frac{\frac{25}{3}}{2}=\frac{25}{6}\)
Vậy \(x=\frac{25}{6};y=\frac{10}{3}\)
b, \(\frac{x}{3}=\frac{y}{4}\)
mà \(xy=192\)
Gọi \(x=3k\)
\(y=4k\)
\(\Rightarrow3k.4k=192\)
\(\Rightarrow12.k^2=192\)
\(\Rightarrow k^2=\frac{192}{12}\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k^2=4^2\)
\(\Rightarrow k=4\)
\(\Rightarrow x=3k=3.4=12\)
\(\Rightarrow y=4k=4.4=16\)
Vậy \(x=12;y=16\)
Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=k\)
Theo đề bài, ta có :
\(xy=54\Rightarrow2k.3k=54\)
\(\Rightarrow5k=54\Rightarrow k=10,8\)
Ta thấy :
\(\dfrac{x}{2}=10,8\Rightarrow x=10,8.2=21,6\)
\(\dfrac{y}{3}=10,8\Rightarrow y=10,8.3=32,4\)
Đặt :\(\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
mà \(xy=54\)
hay 2k . 3k = 54
\(\Rightarrow6.k^2=54\)
\(\Rightarrow k^2=9=\left(\pm3\right)^2\)
Với k = 3 \(\Rightarrow\) \(x=2.3=6;y=3.3=9\)
Với k = -3 \(\Rightarrow x=2.\left(-3\right)=-6;y=3.\left(-3\right)=-9\)
Đặt x = 4k
y = 7k
=> 4k.7k = 112
=> 28.k^2 = 112
=> k^2 = 112 : 28 = 4
=> k = 2
=> x = 4.2 = 8
y = 7.2 = 14
\(\frac{2}{x}=\frac{3}{y}\Rightarrow x=\frac{2y}{3}\)
Thay vào x . y, ta được:
\(x\cdot y=\frac{2y}{3}\cdot y=\frac{2y^2}{3}=96\)
=> \(2y^2=96\cdot3=288\Rightarrow y^2=\frac{288}{2}=144\)
=> \(y=\sqrt{144}=12\) hoặc \(y=-12\)
Vậy x = 8; y = 12 hoặc x = -8 ; y = -12
\(\frac{2}{x}=\frac{3}{y}=>\frac{2}{x}.\frac{3}{y}=\frac{3}{y}.\frac{3}{y}=>\frac{6}{xy}=\frac{9}{y^2}=>\frac{6}{96}=\frac{9}{y^2}=>\frac{1}{16}=\frac{9}{y^2}\)
\(=>y^2=9:\frac{1}{16}=144=12^2=\left(-12\right)^2\)
=>y=12,-12
Với y=12=>x=96:12=8
Với y=-12=>x=96:(-12)=-8
Vậy x=-8,y=-12
x=8,y=12
Câu 1:
a: \(\Leftrightarrow2x^2-x-5< x^2+x-6\)
\(\Leftrightarrow x^2-2x+1< 0\)
hay \(x\in\varnothing\)
b: \(\Leftrightarrow x^2-5x-x+4>0\)
\(\Leftrightarrow x^2-6x+4>0\)
\(\Leftrightarrow\left(x-3\right)^2>5\)
hay \(\left[{}\begin{matrix}x>\sqrt{5}+3\\x< -\sqrt{5}+3\end{matrix}\right.\)
x/2 = y/5
=> xy/10 = x/2 = y/5 = 10/10 = 1
=> x = 1x 2 = 2
y = 1 x 5 = 5
Đặt \(k=\frac{x}{2}=\frac{y}{5}\)
=> \(k^2=\frac{xy}{2.5}=\frac{xy}{10}=\frac{10}{100}=1\)
=> k = -1;1
+ k = -1 thì \(\frac{x}{2}=-1\Rightarrow x=-2\)
\(\frac{y}{5}=-1\Rightarrow y=-5\)
+ k = 1 thi \(\frac{x}{2}=1\Rightarrow x=2\)
\(\frac{y}{5}=1\Rightarrow y=5\)
Vậy .............................
\(\frac{x}{3}=\frac{y}{4}\)mà \(xy=192\)
Ta có \(x=3k\)
\(y=4k\)
\(\Rightarrow3k.4k=192\)
\(\Rightarrow12.k^2=192\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4\)
\(\Rightarrow x=3k=3.4=12\)
\(\Rightarrow y=4k=4.4=16\)
Vậy \(x=12;y=16\)