A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

nhìu giữ cha !!!!

a.

$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.

$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.

$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$

d.

$x^3-3x^2+3x-1=(x-1)^3$

e.

$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$

$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$

f.

$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$

g.

$x^2+2x+1=(x+1)^2$

h. Không phân tích được thành nhân tử

i.

$x^2-2x-3=(x^2-3x)+(x-3)=x(x-3)+(x-3)=(x+1)(x-3)$

j.

$x^2+4x-12=(x^2-2x)+(6x-12)=x(x-2)+6(x-2)=(x-2)(x+6)$

k.

$x^2-8x-9=(x^2+x)-(9x+9)=x(x+1)-9(x+1)=(x+1)(x-9)$

l.

$x^2+x-6=(x^2+3x)-(2x+6)=x(x+3)-2(x+3)=(x-2)(x+3)$

a) x³+4x-5 = x³-x²+x²+4x-5=(x³-x²)+(x²-x)+(5x-5)=x²(x-1)+x(x-1)+5(x-1)=(x²+x+5)(x-1)

b) x³-3x²+4=x³-2x²-x²+4=(x³-2x²)-(x²-4)=x²(x-2)-(x-2)(x+2)=(x²-x+2)(x-2)

c) x³+2x²+3x+2=x³+x²+x²+x+2x+2=(x³+x²)+(x²+x)+(2x+2)=x²(x+1)+x(x+1)+2(x+1)=(x²+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x²+3x)²-2(x²+3x)-8=x⁴+6x³+9x²-2x²-6x-8=x⁴+6x³+7x²-6x-8=x⁴-x³+7x³-7x²+14x²-14x+8x-8=(x⁴-x³)+(7x³-7x²)+(14x²-14x)+(8x-8)=x³(x-1)+7x²(x-1)+14x(x-1)+8(x-1)=(x³+7x²+14x+8)(x-1)=(x³+x²+6x²+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x²+4x+10)²-7(x²+4x+11)+7=(x²+4x+10)²-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

\(=x^3-x+5x-5\)

\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)

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

b) Ta có: \(x^3-3x^2+4\)

\(=x^3+x^2-4x^2+4\)

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

\(=x^3+x^2+x^2+x+2x+2\)

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)

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

e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)

\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)

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

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

\(=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)

\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)

\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)

\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)

Yêu cầu đề là gì vậy bạn?

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

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

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

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

Vậy \(S=\left\{2;2\right\}\)

A= 2x^2 + y^2 - 2xy -2x+3

A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2

A= (x-y)^2 + (x-1)^2 + 2

(x-y)^2> hoặc = 0 với mọi giá trị của x

(x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2

=> A lớn hơn hoặc bằng 2

=> GTNN của A=2 tại x=y=1

+ 2x2 + y2 – 8x + 2y – 1 = 0 không phải phương trình đường tròn vì hệ số của x2 khác hệ số của y2.

+ Phương trình x2 + y2 + 2x – 4y – 4 = 0 có :

a = –1; b = 2; c = –4 ⇒ a2 + b2 – c = 9 > 0

⇒ phương trình trên là phương trình đường tròn.

+ Phương trình x2 + y2 – 2x – 6y + 20 = 0 có :

a = 1; b = 3; c = 20 ⇒ a2 + b2 – c = –10 < 0

⇒ phương trình trên không là phương trình đường tròn.

+ Phương trình x2 + y2 + 6x + 2y + 10 = 0 có :

a = –3; b = –1; c = 10 ⇒ a2 + b2 – c = 0 = 0

⇒ phương trình trên không là phương trình đường tròn.