a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)

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

d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(\Leftrightarrow x^4-4x^3+12x^2-32x+32=\left(y-5\right)^2\)

\(\Leftrightarrow\left(x-2\right)^2\left(x^2+8\right)=\left(y-5\right)^2\)

- Với \(x=2\Rightarrow y=5\)

- Với \(x\ne2\Rightarrow x-2\) là ước của \(y-5\) 

Đặt \(y-5=n\left(x-2\right)\)

\(\Rightarrow\left(x-2\right)^2\left(x^2+8\right)=n^2\left(x-2\right)^2\)

\(\Rightarrow x^2+8=n^2\)

\(\Rightarrow\left(n-x\right)\left(n+x\right)=8\)

\(\Rightarrow\left[{}\begin{matrix}x=1;n=-3\Rightarrow y=8\\x=-1;n=-3\Rightarrow y=14\\x=1;n=3\Rightarrow y=2\\x=-1;n=3\Rightarrow y=-4\end{matrix}\right.\) 

PT có 2 nghiệm

`<=>Delta'>=0`

`<=>4-m^2-1>=0`

`<=>3-m^2>=0`

`<=>m^2<=3`

`<=>-sqrt3<=m<=sqrt3`

Áp dụng vi-ét ta có:`x_1+x_2=4,x_1.x_2=m^2+1`

`3x_1=x_2=>x_1+x_2=4`

`<=>3x_1+x_1=4`

`<=>4x_1=4<=>x_1=1`

`<=>x_2=3`

Mà `m^2+1=x_1.x_2`

`=>m^2+1=3`

`=>m^2=2<=>m=+-sqrt2(tm)`

Vậy `m=+-sqrt2` thì..

Đặt \(\dfrac{1}{x+1}\) = a; \(\dfrac{1}{y}\) = b (x \(\ne\) -1; y \(\ne\) 0)

Khi đó hpt trên tương đương:

\(\left\{{}\begin{matrix}a+b=\dfrac{-1}{2}\\8a+9b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=-4\\8a+9b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-b=1\\8a+9b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=-1\\8a+9\left(-1\right)=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=-1\\8a=4\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=-1\\a=\dfrac{1}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\) (TM)

Vậy hpt có nghiệm duy nhất (x; y) = (1; -1)

Chúc bn học tốt!

ĐK:  ( x ≠ 1 ; y ≠ 0 ) 

Đặt a = \(\dfrac{1}{x+1} \) ; b = \(\dfrac{1}{y}\) . Ta có hệ phương trình 

\(\begin{cases} a + b = \dfrac{-1}{2}\\ 8a + 9b = -5 \end{cases} \)

⇔\(\begin{cases} 8a + 8b = -4 \\ 8a + 9b = -5 \end{cases} \) ⇔ \(\begin{cases} -b = 1 \\ a + b = \dfrac{-1}{2} \end{cases} \) ⇔ \(\begin{cases} b = - 1 \\ a = \dfrac{1}{2} \end{cases} \)

=> \(​​​​\begin{cases} \dfrac{1}{y}=-1 \\\dfrac{1}{x+1}= \dfrac{1}{2} \end{cases} \) ⇔ \(\begin{cases} y = - 1\\ x = 1 \end{cases} \)

Vậy hpt có nghiệm duy nhất \(\begin{cases} y = - 1\\ x = 1 \end{cases} \)

Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{1}{y}=\dfrac{-1}{2}\\\dfrac{8}{x+1}+\dfrac{9}{y}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+1}+\dfrac{8}{y}=-4\\\dfrac{8}{x+1}+\dfrac{9}{y}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-1}{y}=1\\\dfrac{1}{x+1}+\dfrac{1}{y}=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(1;-1)

a) x² + y² + 4x - 10y + 29 = 0

<=> (x² + 4x + 4) + (y² - 10y + 25) = 0

<=> (x+2)² + (y-5)² = 0

Mà: (x+2)² ≥ 0 với mọi x

(y-5)² ≥ 0 với mọi y

=>\(\left\{{}\begin{matrix}\left(x+2\right)^2=0\\\left(y-5\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\y-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)(T/m)

Vậy x = -2 và y = 5.

b) C = 5x² - 20x + 15

= 5(x² - 4x + 3)

= 5(x² - x - 3x + 3)

= 5[x(x-1) - 3(x-1)]

= 5(x-1)(x-3)

c) x² + y² + 2x - 6y + 10 = 0

<=> (x² + 2x + 1) + (y² - 6y + 9) = 0

<=> (x+1)² + (y-3)² = 0

Mà: (x+1)² ≥ 0 với mọi x

(y-3)² ≥ 0 với mọi y

\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)(T/m)

Vậy x = -1 và y = 3

d) A = 3x² - 12x + 15

= 3(x² - 4x + 5)

= 3(x² - 5x + x - 5)

= 3[x(x-5) + (x-5)]

= 3(x-5)(x+1)

đáp án

ko biết

hok tốt

e chưa đến tầm đó