2x+y-5=0
y+x\(^2\)=4x
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26 tháng 8 2018
a. Ta có: x2+y2-2x+4y+5=0
⇌(x-1)2+(y-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b. Ta có: 4x2+y2-4x-6y+10=0
⇌ (2x-1)2+(y-3)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
c.Ta có: 5x2-4xy+y2-4x+4=0
⇌(2x-y)2+(x-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)
d.Ta có: 2x2-4xy+4y2-10x+25=0
⇌ (x-2y)2+(x-5)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)
14 tháng 10 2017
a)
\(A=3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)\(2A=\left[\left(x-y\right)-\left(x+y\right)\right]^2+5\left(x-y\right)^2-5\left(x+y\right)^2\)
\(2A=4y^2+5\left[\left(x-y\right)-\left(x+y\right)\right]\left[\left(x-y\right)+\left(x+y\right)\right]\)\(2A=4y^2+5\left[-2y\right]\left[2x\right]=4y^2-20xy=4y\left(y-5x\right)\\ \)\(A=2y\left(y-5x\right)\)
NM
1
10 tháng 6 2018
1. | x + 1| + (y + 2)2 = 0
Mà (y + 2)2 \(\ge\) 0
Đẳng thức khi . y + 2 \(\ge\) 0
y \(\ge\) - 2
. x + 1 = 0
. x = -1
GH
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
\(\Leftrightarrow2x^2y+y=4x^2+5\)
\(\Leftrightarrow y\left(2x^2+1\right)=4x^2+5\)
\(\Leftrightarrow y=\dfrac{4x^2+5}{2x^2+1}=2+\dfrac{3}{2x^2+1}\)
y nguyên \(\Rightarrow\dfrac{3}{2x^2+1}\) nguyên \(\Rightarrow2x^2+1=Ư\left(3\right)\)
Mà \(2x^2+1\ge1\Rightarrow\left[{}\begin{matrix}2x^2+1=1\\2x^2+1=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=5\\x=1\Rightarrow y=3\\x=-1\Rightarrow y=3\end{matrix}\right.\)
21 tháng 8 2018
a ) \(x^2\left(x+3\right)+y^2\left(y+5\right)-\left(x+y\right)\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow x^3+3x^2+y^3+5y^2-\left(x^3+y^3\right)=0\)
\(\Leftrightarrow3x^2+5y^2=0\)
Do \(\left\{{}\begin{matrix}3x^2\ge0\forall x\\5y^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow3x^2+5y^2\ge0\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2=0\\5y^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(x=0;y=0\)
b )\(\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(-16\left(x^3-y\right)=32\)
\(\Leftrightarrow\left[\left(2x\right)^3-y^3\right]+\left[\left(2x\right)^3+y^3\right]-16x^3+16y=32\)
\(\Leftrightarrow8x^3-y^3+8x^3+y^3-16x^3+16y=32\)
\(\Leftrightarrow16y=32\)
\(\Leftrightarrow y=2\)
Vậy \(y=2\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\y+x^2=4x\end{matrix}\right.\)
\(\Rightarrow5-2x+x^2=4x\)
\(\Leftrightarrow x^2-6x+5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Nếu \(x=1\Rightarrow y=5-2x=3\)
Nếu \(x=5\Rightarrow y=5-2x=-5\)
Vậy hpt đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(5;-5\right)\right\}\)