\(C=4x^2+10y-4x+10y-2\)

\(=\left(4x^2-4x+1\right)+\left(10y^2+10y+\frac{5}{2}\right)-\frac{11}{2}\)

\(=\left(2x-1\right)^2+\left(\sqrt{10y}+\sqrt{\frac{5}{2}}\right)^2-\frac{11}{2}\ge\frac{-11}{2}\)

Vậy \(C_{min}=-\frac{11}{2}\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

và \(\sqrt{10}y+\sqrt{\frac{5}{2}}=0\Leftrightarrow y\frac{-\sqrt{5}}{\sqrt{20}}=-0,5\)

=>x^3+2x^2+2x^2+4x-5x-10+7 chia hết cho x+2

=>\(x+2\in\left\{1;-1;7;-7\right\}\)

=>\(x\in\left\{-1;-3;5;-9\right\}\)

mik cần gấp giúp vs ạ

\(2x^2+9y^2+6xy-18y-8x+15\)

\(=\left(x^2+6xy+9y^2\right)-6x-18y+9+\left(x^2-2x+1\right)+5\)

\(=\left(x+3y\right)^2-6\left(x+3y\right)+9+\left(x-1\right)^2+5\)

\(=\left(x+3y-3\right)^2+\left(x-1\right)^2+5\)

\(\ge5\)

Dấu "=" xảy ra tại \(x=1;y=\frac{2}{3}\)

Vậy......

a, = [(x-2).(x+1)]^2+(x-2)^2

    = (x-2)^2.(x+1)^2+(x-2)^2

    = (x-2)^2.[(x+1)^2+1]

    = (x-2)^2.(x^2+2x+2)

Tk mk nha

b)  \(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

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

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

3x(x+5)-2x-10=0

<=>3x(x+5)-(2x+10)=0

<=>3x(x+5)-2(x+5)=0

<=>(3x-2)(x+5)=0

<=>\(\hept{\begin{cases}3x-2=0\\x+5=0\end{cases}}\)<=>\(\hept{\begin{cases}x=\frac{2}{3}\\x=\left(-5\right)\end{cases}}\)

vậy tập nghiệm cua phương trình là S={\(\frac{2}{3};-5\)}

cám ơn

b, x² +y²+z² +2x-4y-6z+14=0

<=> (x²+2x+1)+(y²-4y+4)+(z²-6z+9)=0

<=> (x+1)²+(y-2)²+(z-3)²=0

=>(x+1)²=(y-2)²=(z-3)²=0

=>x+1=y-2=z-3=0

=> x=-1; y=2; z=3

c, 2x²+y²-6x-4y+2xy+5=0

<=> (x²+y²+4+2xy-4x-4y)+(x²-2x+1)=0

<=> (x+y-2)²+(x-1)²=0

=> (x+y-2)²=(x-1)²=0

=>x+y-2=x-1=0

=>x=1; y=1