x/y=5/3 và x-y=20. tìm x,y
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Câu 3:
\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)
\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)
Câu b:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)
\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)
Câu c:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)
\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)
Câu d:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)
\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)
Câu e:
\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)
\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)
3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)
4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)
\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)
5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)
6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)
7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)
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\(\dfrac{4}{3}\) = \(\dfrac{20}{4}\)\(x\) - 1
5\(x\) - 1 = \(\dfrac{4}{3}\)
5\(x\) = \(\dfrac{4}{3}\) + 1
5\(x\) = \(\dfrac{7}{3}\)
\(x\) = \(\dfrac{7}{3}\) : 5
\(x\) = \(\dfrac{7}{15}\)
b,
2\(\times\)\(x\) = 3\(\times\) y
\(x\) = \(\dfrac{3}{2}\) \(\times\) y
y - \(\dfrac{3}{2}\) \(\times\) y = 5
\(y\) \(\times\) ( 1 - \(\dfrac{3}{2}\)) = 5
\(y\) \(\times\) - \(\dfrac{1}{2}\) = 5
\(y\) = 5 : (-\(\dfrac{1}{2}\))
\(y\) = - 10
\(x\) = y - 5 = -10 - 5 =-15
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Ta có \(\frac{x}{3}=\frac{y}{2};\frac{y}{3}=\frac{z}{5}\Leftrightarrow\frac{x}{9}=\frac{y}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{9}=\frac{y}{6}=\frac{z}{10}=\frac{x-y+z}{9-6+10}=\frac{20}{13}\)
Từ \(\frac{x}{9}=\frac{20}{13}\Rightarrow x=\frac{20.9}{13}=\frac{180}{13}\)
\(\frac{y}{6}=\frac{20}{13}\Rightarrow y=\frac{20.6}{13}=\frac{120}{13}\)
\(\frac{z}{10}=\frac{20}{13}\Rightarrow z=\frac{20.10}{13}=\frac{200}{13}\)
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a) 2x+1.3y=123
<=>2x+1.3y=(22)3.33
<=> 2x+1=26 và 3y=33
<=>x+1=6 và y=3
<=>x=5 và y=3
b) 10x : 5y=20y
<=>10x=20y.5y=100y=(102)y
<=>x=2y (Nhiều số lắm chèn)
c) 2x=4y-1
<=>2x=2y-2
<=>x=y-2
Mặt khác: 27y=3x+8
<=> 33y=3x+8
<=>3y=x+8
<=>3y=(y-2)+8
<=>2y=6
<=>y=3
=>x=y-2=3-2=1
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\(\frac{x}{y}=\frac{20}{3}\Rightarrow x=\frac{20y}{3}\) \(\Rightarrow xy=\frac{20y}{3}.y=\frac{27}{5}\Leftrightarrow y^2=\frac{27}{5}.\frac{3}{20}=\frac{81}{100}\)
\(\Rightarrow y=\frac{9}{10}\)hoặc \(y=-\frac{9}{10}\)
Với \(y=\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{10}{9}=\frac{270}{45}=6\)
Với \(y=-\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{-10}{9}=-6\)
Vậy \(\left(x;y\right)\in\left\{\left(6;\frac{9}{10}\right);\left(-6;-\frac{9}{10}\right)\right\}.\)
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\(\text{a)}\)\(2^{x+1}.3^y=2^{2x}.3^x\Leftrightarrow\frac{2^{2x}}{2^{x+1}}=\frac{3^y}{3^x}\)
\(\Leftrightarrow2^{x-1}=3^{y-x}\)
\(\Leftrightarrow x-1=y-x=0\)
\(\Leftrightarrow x=y=1\)
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Giải:
Ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\)
\(\frac{y}{2}=\frac{z}{5}\Rightarrow\frac{y}{6}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{-20}{25}=\frac{-4}{5}\)
+) \(\frac{x}{4}=\frac{-4}{5}\Rightarrow x=\frac{-16}{5}\)
+) \(\frac{y}{6}=\frac{-4}{5}\Rightarrow y=\frac{-24}{5}\)
+) \(\frac{z}{15}=\frac{-4}{5}\Rightarrow z=-12\)
Vậy bộ số \(\left(x,y,z\right)\) là \(\left(\frac{-16}{5},\frac{-24}{5},-12\right)\)
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a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)
=>x=-3; y=28; z=-60
b: 5/12=x/-72
=>x=-72*5/12=-6*5=-30
c: =>x+3=-5
=>x=-8
Ta có:
\(\frac{x}{y}=\frac{5}{3}\Rightarrow\frac{x}{5}=\frac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{3}=\frac{x-y}{5-3}=\frac{20}{2}=10\)
\(\frac{x}{5}=10\Rightarrow x=50\)
\(\frac{y}{3}=10\Rightarrow y=30\)
hiệu của x và y ứng với : 5-3 =2 ( phần )
-> 1 phần bằng 20 :2 =10
-> x =10 *3 = 30
-> y= 30 + 20 =50