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

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

\(-18x+13=0\)

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

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

b: \(\left(4x-1\right)\left(2x^2-x-1\right)\)

\(=8x^3-4x^2-4x-2x^2+x+1\)

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

\(-2x^2-2xy-y^2+2x-2y-2=-\left[y^2+2y\left(x+1\right)+\left(x+1\right)^2\right]-\left(x^2-4x+4\right)+3=-\left(y+x+1\right)^2-\left(x-2\right)^2+3\le3\)

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

XL gtnn B = 19/4

GTNN = -1/4

GTNN B = 23/4

`A=2x^2-2xy-6x+y^2+10`

`A=x^2-2xy+y^2+x^2-6x+10`

`A=(x-y)^2+x^2-6x+9+1`

`A=(x-y)^2+(x-3)^2+1`

Vì `(x-y)^2+(x-3)^2>=0=>A>=1`

Dấu "=" xảy ra khi `{(x-y=0),(x-3=0):}<=>x=y=3`

A=\(\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\\ \)

dấu= xảy ra khi x=y=3

tick mik nha

c)    \(2xy-x^2-y^2+16\)

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

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

\(=\left(4-x+y\right)\left(4+x-y\right)\)

c ) \(2xy - x^2 - y^2 + 16\)

 \(= 16 - ( x^2 - 2xy + y^2 ) \)

\(= 16 - ( x - y ) ^2 \)

\(= ( 4 - x + y )\)

\(( 4 + x - y )\)

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

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

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

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

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

\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+4\\ A=\left(x-y\right)^2+\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\Leftrightarrow x=y=1\)