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1.
\(x^2\)+\(y^2\)+2y-6x+10=0
=> \(x^2\)-6x+9 +\(y^2\)+2y+1=0
=> (x-3)\(^2\)+(y+1)\(^2\)=0
pt vô nghiệm
4.
=> \(x^2\)+8x+16+(3y)\(^2\)-2.3.2y+4=0
=> (x+4)\(^2\)+(3y-2)\(^2\)=0
pt vô nghiệm
1) \(4x^2+4x+6y+9y^2+2=0\Leftrightarrow\left(4x^2+4x+1\right)+\left(9y^2+6y+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)^2+\left(3y+1\right)^2=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-1}{3}\end{matrix}\right.\)
vậy \(x=\dfrac{-1}{2};y=\dfrac{-1}{3}\)
2) \(25x^2+9y^2-10x+12y+5=0\Leftrightarrow\left(25x^2-10x+1\right)+\left(9y^2+12y+4\right)=0\)
\(\Leftrightarrow\left(5x-1\right)^2+\left(3y+2\right)^2=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(5x-1\right)^2=0\\\left(3y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-1=0\\3y+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=1\\3y=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{-2}{3}\end{matrix}\right.\)
vậy \(x=\dfrac{1}{5};y=\dfrac{-2}{3}\)
3) \(9x^2+4y^2+12x-8y+17=0\Leftrightarrow\left(9x^2+12x+4\right)+\left(4y^2-8y+4\right)+9=0\)
\(\Leftrightarrow\left(3x+2\right)^2+\left(2y-2\right)^2+9=0\)
ta có : \(\left(3x+2\right)^2\ge0\forall x\) và \(\left(2y-2\right)^2\ge0\forall y\)
\(\Rightarrow\) \(\left(3x+2\right)^2+\left(2y-2\right)^2+9\ge9>0\forall x;y\)
\(\Rightarrow\) phương trình vô nghiệm
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
1) \(9x^2+y^2-10y-12x+29=0\)
\(\Leftrightarrow\left(9x^2-12x+4\right)+\left(y^2-10y+25\right)=0\)
\(\Leftrightarrow\left(3x-2\right)^2+\left(y-5\right)^2=0\)
ta có : \(\left(3x-2\right)^2\ge0\forall x\) và \(\left(y-5\right)^2\ge0\forall y\)
\(\Rightarrow\left(3x-2\right)^2+\left(y-5\right)^2=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2\right)^2=0\\\left(y-5\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=5\end{matrix}\right.\)
vậy \(x=\dfrac{2}{3};y=5\)
2) câu này đề sai rồi nha
3) \(x^2+29+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)+9=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9=0\)
ta có : \(\left(x+4\right)^2\ge0\forall x\) và \(\left(3y-2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9\ge9>0\forall x;y\)
vậy phương trình vô nghiệm
1. Ta có:
\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)
Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)
2.
a) \(x^2+4y^2+6x-12y+18=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)
Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)
Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)
b) \(2x^2+2y^2+2xy-10x-8y+41=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)
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Rồi giải tương tự như trên
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) \(x^2+26+6y+9y^2-10x\)
\(=\left(x^2-10x+25\right)+\left(9y^2+6y+1\right)\)
\(=\left(x-5\right)^2+\left(3y+1\right)^2\)
Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
