a) x² + y² + 2x - 4y + 5 = 0

 <=> ( x² + 2x +1 ) + ( y² - 4y + 4 ) = 0

 <=> ( x + 1 )² + ( y - 2 ) ² = 0

 <=> \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)

 <=> \(\hept{\begin{cases}x+1=0\\y-2=0\end{cases}}\)

 <=> \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

   b) x² + 4y² - x + 4y + \(\frac{5}{4}\)=0

<=> ( x² - 2x + \(\frac{1}{4}\)) + ( 4y² + 4y + 1 ) = 0

<=> ( x - \(\frac{1}{2}\))²  + ( 2y + 1 )² = 0

<=> \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\2y+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\2y=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-1}{2}\end{cases}}\)

a) Ta có \(x^2+y^2+2x-4y+5=0\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)

<=> x=-1;y=2

b)Ta có:\(x^2+4y^2-x+4y+\frac{5}{4}=0\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+1\right)=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)

<=> x=1/2 ;y=-1/2

a, \(x^2+y^2+2x-4y+5=0\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0.\)

    \(\left(x+1\right)^2+\left(y-2\right)^2=0\)

   \(\Rightarrow x+1=0\)và \(y-2=0\)

\(\left(+\right)x+1=0\Rightarrow x=-1\)

\(\left(+\right)y-2=0\Rightarrow y=2\)

Vậy x=-1 ; y=2 

b, \(x^2+4y^2-x+4y+\frac{5}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+\frac{4}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\) và \(2y+1=0\)

\(\left(+\right)x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(\left(+\right)2y+1=0\Rightarrow2y=-1\Rightarrow y=-\frac{1}{2}\)

Vậy \(x=\frac{1}{2};y=-\frac{1}{2}\)

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+y^4-4xy^3=1\\4x^2+2y^2-4xy=2\end{matrix}\right.\)

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

b) 4x^2+y^2-20x-2y+26=0;
(4x^2-20x+25)+(y^2-2y+1)=(2x-5)^2+(y-1)^2=0
<=>x=5/2; y=1

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2y+4x^2+6y-30=0\\x^4+y^2-2x^2-4y-5=0\end{matrix}\right.\)

\(\Rightarrow x^4+2x^2y+y^2+2x^2+2y-35=0\)

\(\Rightarrow\left(x^2+y+1\right)^2-36=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+y+1=6\\x^2+y+1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=5-y\\x^2=-7-y\end{matrix}\right.\)

Thế vào pt đầu ...

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)

1.

Đường tròn tâm \(I\left(0;0\right)\) bán kính \(R=1\)

\(d\left(I;A\right)=\frac{\left|3.0-4.0+5\right|}{\sqrt{3^2+\left(-4\right)^2}}=\frac{5}{5}=1=R\)

\(\Rightarrow\) Đáp án A đúng

2.

Do d vuông góc \(2x-y+4=0\) nên d nhận \(\left(1;2\right)\) là 1 vtpt

Phương trình d:

\(1\left(x+1\right)+2\left(y-2\right)=0\Leftrightarrow x+2y-3=0\)