y x 2 + y/2 = 10. tìm y
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A) \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2.4}=\frac{y}{3.4}\Rightarrow\frac{x}{8}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{3.4}=\frac{z}{3.5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)
Từ 1 và 2
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)s
Áp dụng tính chất dãy tỉ số bằng nhau
=> \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\frac{x}{8}=2\Rightarrow x=16\)
\(\frac{y}{12}=2\Rightarrow y=24\)
\(\frac{z}{15}=2\Rightarrow z=30\)
B) Đặt \(\frac{x}{2}=\frac{y}{5}=k\)
=> \(\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)
xy = 10
=> 2k . 5k = 10
=> 10 . k2 = 10
=> k2 = 1
=> \(\hept{\begin{cases}k=-1\\k=1\end{cases}}\)
=> Với \(\hept{\begin{cases}k=-1\Rightarrow\hept{\begin{cases}x=-2\\y=-5\end{cases}}\\k=1\Rightarrow\hept{\begin{cases}x=2\\y=5\hept{\begin{cases}\\\end{cases}}\end{cases}}\end{cases}}\)
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a) 2y - 12y = 0
\(\Rightarrow\) y ( 2-12) = 0
\(\Rightarrow\) y . (-10) =0
\(\Rightarrow\) y = 0 : (-10) = 0
b) (y-7)(y-8) = 0
\(\Rightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}\Rightarrow}\orbr{\begin{cases}y=7\\y=8\end{cases}}}\)
c) x + x.2+x.3+x.4+...+x.10 = 165
\(\Rightarrow\) x ( 1+2+3+.....+8+9+10) = 165
\(\Rightarrow\)x . \(\frac{\left(1+10\right).10}{2}\)=165
\(\Rightarrow\) x . 55 = 165
\(\Rightarrow x=\frac{165}{55}=3\)
Can you k for me ,Lê Thị Kim Chi!
a) \(2y-12y=0\)
\(\Leftrightarrow-10y=0\)
\(\Leftrightarrow y=0:\left(-10\right)\)
\(\Leftrightarrow y=0\)
b) \(\left(y-7\right)\left(y-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=7\\y=8\end{cases}}\)
c) \(x+x.2+x.3+......+x.10=165\)
\(\Leftrightarrow x.\left(1+2+3+.....+10\right)=165\)
\(\Leftrightarrow x.55=165\)
\(\Leftrightarrow x=165:55\)
\(\Leftrightarrow x=3\)
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\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{y^2+x^2+y^2-x^2}{3+5}=\frac{2y^2}{8}=\frac{y^2}{4}\)
\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{\left(x^2+y^2\right)-\left(y^2-x^2\right)}{5-3}=\frac{2x^2}{2}=x^2\)
\(\frac{y^2}{4}=x^2\Rightarrow\frac{y^{10}}{1024}=\frac{x^{10}}{1}\Rightarrow x^{20}=\frac{x^{10}.y^{10}}{1024}=\frac{1024}{1024}=1\)
=>x=-1;1
xét x=-1=>y2=4=>y=-2;2
xét x=1=>y2=4=>y=-2;2
Vậy (x;y)=(-1;-2);(-1;2);(1;-2);(1;2)
\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{y^2+x^2+y^2-x^2}{3+5}=\frac{2y^2}{8}=\frac{y^2}{4}\)
\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}=\frac{\left(x^2+y^2\right)-\left(y^2-x^2\right)}{5-3}=\frac{2x^2}{2}=x^2\)
\(\frac{y^2}{4}=x^2\Rightarrow\frac{y^{10}}{1024}=\frac{x^{10}}{1}\Rightarrow x^{20}=\frac{x^{10}.y^{10}}{1024}=\frac{1024}{1024}=1\)
=>x=-1;1
xét x=-1=>y2=4=>y=-2;2
xét x=1=>y2=4=>y=-2;2
Vậy (x;y)=(-1;-2);(-1;2);(1;-2);(1;2)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất của dãy tỉ số bằng nhau có: \(\frac{y^2-x^2}{3}=\frac{x^2+y^2}{5}=\frac{\left(y^2-x^2\right)+\left(x^2+y^2\right)}{3+5}=\frac{\left(y^2-x^2\right)-\left(x^2-y^2\right)}{3-5}\)
=> \(\frac{2y^2}{8}=\frac{-2x^2}{-2}\Rightarrow\frac{y^2}{4}=x^2\) => y2 = 4x2
Ta có x10.y10 = x10. (4x2)5 = 1024.x20 = 1024 => x20 = 1 => x =1 hoặc x = -1
=> y2 = 4 => y = 2 hoặc y = -2
Vậy ...
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y*2+y*1/2=10
y*(2+1/2)=10
y*(4/2+1/2)=10
y*5/2=10
y=10:5/2=4
Vậy y=4
Ta có: y x 2 + y : 2 = 10
=> y x 2 + y x 1/2 = 10
=> y x (2 + 1/2) = 10
=> y x 5/2 = 10
=> y = 20 : 5/2
=> y = 4