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a/
\(\Leftrightarrow\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)-19=0\)
\(\Leftrightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2=19\)
Do 19 không thể phân tích thành tổng của 2 số chính phương nên pt vô nghiệm
b/
\(\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do x; y nguyên dương nên \(\left(2x+2y\right)^2>0\Rightarrow VT>0\)
Pt vô nghiệm
c/
\(\Leftrightarrow\left(x^2+4y^2+25-4xy+10x-20y+25\right)+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)
Do x;y;z nguyên dương nên \(\left|x+y+z\right|>0\Rightarrow VT>0\)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Do x;y;z nguyên dương nên vế phái luôn dương
Pt vô nghiệm
Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
a: \(x^2-10x+26+y^2+2y=0\)
\(\Leftrightarrow x^2-10x+25+y^2+2y+1=0\)
\(\Leftrightarrow\left(x-5\right)^2+\left(y+1\right)^2=0\)
=>x=5 hoặc y=-1
b: \(x^2-6x+13+y^2+4y=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=0\)
=>x=3 và y=-2
a) \(\Rightarrow\left(x^2+2\times5x+25\right)+\left(y^2+2y+1\right)\)
\(\Rightarrow\left(x+5\right)^2+\left(y+1\right)^2\)
Bài 1:
a) \(x^2+10x+26+y^2+2y\)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(4x^2-y^2-12x+2y+8\)
\(=4x^2-12x+9-y^2+2y-1\)
\(=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
Bài 2:
\(P=4+8x-16x^2\)
\(P=-\left(16x^2-8x+4\right)\)
\(P=-\left[\left(4x\right)^2-2.4x+1+3\right]\)
\(P=-\left(4x-1\right)^2-3\)
Vì \(-\left(4x-1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(4x-1\right)^2-3\le-3\) với mọi x
\(\Rightarrow Pmax=-3\Leftrightarrow4x-1=0\)
\(\Rightarrow4x=1\)
\(\Rightarrow x=\dfrac{1}{4}\)
Vậy Pmax = -3 <=> x = 1/4
1) \(4x^2-12x+y^2-4y+13\)
\(=\left(4x^2-12x+9\right)+\left(y^2-4y+4\right)\)
\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(y^2-2.2y+4\right)\)
\(=\left(2x-3\right)^2+\left(y-2\right)^2\)
2) \(x^2+y^2+2y-6x+10\)
\(=\left(x^2+2y+1\right)+\left(y^2-6x+9\right)\)
\(=\left(x+1\right)^2+\left(y-3\right)^2\)
3) \(4x^2+9y^2-4x+6y+2\)
\(=\left(4x^2-4x+1\right)+\left(9y^2+6y+1\right)\)
\(=\left(2x-1\right)^2+\left(3y+1\right)^2\)
4) \(y^2+2y+5-12x+9x^2\)
\(\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)\)
\(=\left(y+1\right)^2+\left(3x-2\right)^2\)
5) \(x^2+26+6y+9y^2-10x\)
\(=\left(x^2-10x+25\right)+\left(9y^2+6y+1\right)\)
\(=\left(x-5\right)^2+\left(3y+1\right)^2\)
a) \(x^2+10x+26+y^2+2y\)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(x^2-2xy+2y^2+2y+1=\left(x-y\right)^2+\left(y+1\right)^2\)
thay 2014 = x + 1
sau đó biến đổi rút gọn
a) \(x^2+10x+26+y^2+2y\)
\(=\left(x^2+10x+25\right)+\left(1+2y+y^2\right)\)
\(=\left(x+5\right)^2+\left(1+y\right)^2\)
b) \(x^2-2xy+2y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c) \(2x^2+2y^2=2\left(x^2+y^2\right)\)
\(5x^2+2y^2+13+10x+2y\)
\(=5x^2+10x+5+2y^2+2y+\frac{1}{2}+7\frac{1}{2}\)
\(=5\left(x^2+2x+1\right)+2\left(y^2+y+\frac{1}{4}\right)+7\frac{1}{2}\)
\(=5\left(x+1\right)^2+2\left(y+\frac{1}{4}\right)^2+7\frac{1}{2}>0\forall x;y\)
dẫn đến mâu thuẫn so với đề bài.
Vậy \(x,y\in\varnothing\)
Chúc bạn học tốt.
a) \(x^2+10x+26+y^2+2y\)
\(=x^2+2.5x+25+1+y^2+2y\)
\(=\left(x^2+2.5x+25\right)+\left(1+2y+y^2\right)\)
\(=\left(x+5\right)^2+\left(1+y\right)^2\)
b) \(x^2-2xy+2y^2+2y+1\)
\(=x^2-2xy+y^2+y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c) \(z^2-6z+13+t^2+4t\)
\(=z^2-2.3z+9+4+t^2+4t\)
\(=\left(z^2-2.3x+9\right)+\left(4+4t+t^2\right)\)
\(=\left(z-3\right)^2+\left(2+t\right)^2\)
d) \(4x^2+2z^2-4xz-2z+1\)
\(=4x^2+z^2+z^2-4xz-2z+1\)
\(=\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)\)
\(=\left(2x-z\right)^2+\left(z-1\right)^2\)
\(x^2+y^2+26+10x-2y=0\)
\(\left(x^2+10x\right)+\left(y^2-2y\right)+26=0\)
\(\left(x^2+2.x.5+5^2\right)+\left(y^2-2.y.1+1^2\right)=0\)
\(\left(x+5\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\y-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\y=1\end{array}\right.\)
j) \(x^2+y^2+26+10x-2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\begin{cases}x+5=0\\y-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-5\\y=1\end{cases}\)
Vậy x=-5; y=1