2 x y = 120
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y x 2 + y x 4 + y x 2 = 120
y x ( 2 + 4 + 2 ) = 120
y x 10 = 120
y = 120 : 10
y = 12
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y x 2 + y x 4 + y x 2 = 120
y x ( 2 + 4 + 2 ) = 120
y x 10 = 120
y = 120 : 10
y = 12
cong chua nu cuoi
y x 2 + y x 4 + y x 2 = 120
y x ( 2 + 4 + 2 ) = 120
y x 8 = 120
y = 120 : 8
y = 15
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y x 2 + y x 4 + y x 2 = 120
y x ( 2 + 4 + 2 ) = 120
y x 8 = 120
y = 120 : 8
y = 15
y x 2 + y x 4 + y x 2 = 120
y x ( 2 + 4 + 2 ) = 120
y x 10 = 120
y = 120 : 10
y = 12
Le anh tu
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1) ( y-25):4 - 120 = 0 2) 120 - ( x+25 )x4 = 0
( y-25 ):4 = 0+120 (x+25) x4 =120-0
( y-25):4 =120 (x+25) x4 =120
( y-25)=120x4 (x+25)=120:4
y-25=480 x+25=30
y=480+25 x=30-25
y=505 x=5
k mk nha
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y x 2 + y x 4 + y x 6 = 120
y x ( 2+ 4 + 6 ) = 120
y x 8 = 120
y = 120 : 8
y = 15
y x 2 + y x 4 + y x 6 = 120
y x ( 2 + 4 + 6 ) = 120
y x 12 = 120
y = 120 / 12
y = 10
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\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)
Dựa vào tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+y}{2+5}=\dfrac{120}{7}\)
=> \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{120}{7}\)
=> \(\left\{{}\begin{matrix}x=34,28...\\y=85,71...\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{120}{7}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{240}{7}\\y=\dfrac{600}{7}\end{matrix}\right.\)
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\(\left(2x+1\right)\left(y-1\right)=-7\\ \Rightarrow2x+1;y-1\in U_{\left(-7\right)}=\left\{-7;-1;1;7\right\}\)
\(TH1\) | \(TH2\) | \(TH3\) | \(TH4\) | |
\(2x+1\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
\(y-1\) | \(-7\) | \(7\) | \(-1\) | \(1\) |
\(x\) | \(0\) | \(-1\) | \(3\) | \(-4\) |
\(y\) | \(-6\) | \(8\) | \(0\) | \(2\) |
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Do \(x-y=2\Rightarrow\left(x-y\right)^2=4\)
\(\Rightarrow x^2-2xy+y^2=4\)\(\Rightarrow x^2+y^2=4+2xy=4+2.120=244\)
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\(\left\{{}\begin{matrix}x-y=10\\\dfrac{-120\left(x-y\right)}{xy}=\dfrac{2}{5}\end{matrix}\right.\) \(\Rightarrow\dfrac{-1200}{xy}=\dfrac{2}{5}\Rightarrow xy=-3000\)
Ta được hệ: \(\left\{{}\begin{matrix}x-y=10\\xy=-3000\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+10\\xy=-3000\end{matrix}\right.\)
Thay pt trên vào dưới:
\(\left(y+10\right).y=-3000\Rightarrow y^2+10y+3000=0\)
\(\Rightarrow\) pt vô nghiệm
Vậy hệ đã cho vô nghiệm
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\(x^2+y^2=\left(x+y\right)^2-2xy=\left(-8\right)^2-2\cdot15=34\)
Chọn D.
`2 xx y = 120`
`=> y = 120 : 2`
`=> y = 60`
Vậy ...
60