a) \(2x^2-5x^2+6x+13=0\)

\(\Leftrightarrow-3x^2+6x+13=0\)

\(\Leftrightarrow3x^2-6x-13=0\left(1\right)\)

\(\Delta'=9+39=48>0\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{3}\)

Pt (1) có 2 nghiệm phân biệt là :

\(\left[{}\begin{matrix}x=\dfrac{3+4\sqrt[]{3}}{3}=1+\dfrac{4\sqrt[]{3}}{3}\\x=\dfrac{3-4\sqrt[]{3}}{3}=1-\dfrac{4\sqrt[]{3}}{3}\end{matrix}\right.\)

b) \(x^2-5x=-4\)

\(\Leftrightarrow x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)

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

\(\left(5x^2-7x^2y^3+3y^4\right)-K=3x^2-7x^2y^3-3y^4\)

\(\Rightarrow K=\left(5x^2-7x^2y^3+3y^4\right)-\left(3x^2-7x^2y^3-3y^4\right)\)

\(\Rightarrow K=5x^2-7x^2y^3+3y^4-3x^2+7x^2y^3+3y^4\)

\(\Rightarrow K=2x^2+6y^4\)

________________

\(3x^2-8x+5-K=-2K+4x-6+x^2\)

\(\Rightarrow-K+2K=\left(4x-6+x^2\right)-\left(3x^2-8x+5\right)\)

\(\Rightarrow K=4x-6+x^2-3x^2+8x-5\)

\(\Rightarrow K=-2x^2+12x-11\)

Ta co´:

             -4 ( 2x + 9 ) - ( -8x + 3 ) - ( x + 13 ) = 0

                                -8x - 36 + 8x -3 - x - 13 = 0

            .             ( -8x + 8x ) - 36 - 3 - x - 13 = 0

                                              - 36 - 3 - x - 13 = 0

                                                                     x = 0 + 36 + 3 + 13

                                                                     x = 52

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+8\left(x-y\right)+16=3-2y^2\)

\(\Leftrightarrow\left(x-y\right)^2+8\left(x-y\right)+16=3-2y^2\)

\(\Leftrightarrow\left(x-y+4\right)^2=3-2y^2\) (1)

Do \(\left(x-y+4\right)^2\ge0;\forall x,y\)

\(\Rightarrow3-2y^2\ge0\Rightarrow y^2\le\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}y^2=0\\y^2=1\end{matrix}\right.\)

\(\Rightarrow y=\left\{-1;0;1\right\}\)

- Với \(y=-1\) thay vào (1):

\(\left(x+5\right)^2=1\Rightarrow\left[{}\begin{matrix}x+5=1\\x+5=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\)

- Với \(y=1\) thay vào (1):

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

- Với \(y=0\)

\(\Rightarrow\left(x+4\right)^2=3\) (ko có nghiệm nguyên do 3 ko phải SCP)

\(D=2023-8x+2y+4xy-y^2-5x^2\)

\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)

\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)

\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)

\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)

\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)

Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)

\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

a: f(1)=a+b+c=0

=>x=1 là nghiệm

b: Vì 5-6+1=0

nên f(x)=5x^2-6x+1 có một nghiệm là x=1

A=5x2+2y2−4xy−8x−4y+19=(2x2−4xy+2y2)+4(x−y)+(3x2−12x)+19=2(x−y)2+4(x−y)+3(x2−4x+4)+7=2[(x−y)2+2(x−y)+1]+3(x−2)2+5=2(x−y+1)2+3(x−2)2+5≥0Dấu "=" xảy ra khi{x−y+1

mik viết 5x2 là 5x mũ 2 nha

5x² - 15x = 0

5x(x-3)=0

suy ra 2 trường hợp

x=0

x-3=0=>x=3

5x²-15x=0
5x(x-3) =0
TH1: 5x=0            TH2: x-3=0
       =>x=0                   => x=3
    Vậy x thuộc {0;3}