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ta có : \(4x^2+4y^2-2xy-6x-6y+6\)
\(=x^2-2xy+y^2+3x^2-6x+3+3y^2-6y+3\)
\(=\left(x-y\right)^2+3\left(x-1\right)^2+3\left(y-1\right)^2\ge0\forall x;y\left(đpcm\right)\)
Lời giải:
a)
Ta có: \(x^2+10x+30=x^2+2.x.5+5^2+5=(x+5)^2+5\)
Vì $(x+5)^2\geq 0, \forall x\Rightarrow x^2+10x+30=(x+5)^2+5\geq 5>0$ (đpcm)
b)
\(4x-x^2-7=-(x^2-4x+7)=-(x^2+4x+4+3)=-[(x-2)^2+3]\)
Vì $(x-2)^2\geq 0, \forall x\Rightarrow (x-2)^2+3\geq 3>0$
$\Rightarrow 4x-x^2-7=-[(x-2)^2+3]< 0$ (đpcm)
c)
\(x^2+4y^2-2x-4y+2=(x^2-2x+1)+(4y^2-4y+1)\)
\(=(x-1)^2+(2y-1)^2\)
Vì $(x-1)^2\geq 0; (2y-1)^2\geq 0, \forall x,y$
$\Rightarrow x^2+4y^2-2x-4y+2=(x-1)^2+(2y-1)^2\geq 0$ (đpcm)
1)
a) \(2x^2-12x+18+2xy-6y\)
\(=2x^2-6x-6x+18+2xy-6y\)
\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)
\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)
\(=\left(x-3\right)\left(2y+2x-6\right)\)
\(=2\left(x-3\right)\left(y+x-3\right)\)
b) \(x^2+4x-4y^2+8y\)
\(=x^2+4x-4y^2+8y+2xy-2xy\)
\(=\left(-4y^2+2xy+8y\right)+\left(-2xy+x^2+4x\right)\)
\(=2y\left(-2y+x+4\right)+x\left(-2y+x+4\right)\)
\(=\left(2y+x\right)\left(-2y+x+4\right)\)
2) \(5x^3-3x^2+10x-6=0\)
\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)
Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)
\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)
\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
Bài làm
a) 2x2 - 12x + 18 + 2xy - 6y
= 2x2 - 6x - 6x + 18 + 2xy - 6y
= ( 2xy + 2x2 - 6x ) - ( 6y + 6x - 18 )
= 2x( y + x - 3 ) - 6( y + x - 3 )
= ( 2x - 6 ) ( y + x - 3 )
# Học tốt #
\(2x^2+2y^2-2xy-4x-4y+8\)
\(=x^2-2xy+y^2+x^2-4x+y^2-4y+8\)
\(=\left(x-y\right)^2+x^2-4x+4+y^2-4x+4\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\)
\(\RightarrowĐPCM\)
Ta có x2 - 2x + 5
= (x2 - 2x + 4) + 1
= (x - 2)2 + 1 \(\ge\)1 > 0 (đpcm)
b) Ta có : 4x2 + 4x - 3 = (4x2 + 4x + 1) - 4 = (2x + 1)2 - 4 \(\ge\) - 4 (đpcm)
+) Ta có: \(x^2-2x+5=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-1\right)^2+4\ge4>0\forall x\)
Vậy \(x^2-2x+5>0\)
Phân tích đa thức thành nhân tử
a. \(12x^3y-24x^2y^2+12xy^3\)
\(=12xy\left(x^2-2xy+y^2\right)\)
\(=12xy\left(x-y\right)^2\)
b. \(x^2-6x+xy-6y\)
\(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x+y\right)\left(x-6\right)\)
c. \(2x^2+2xy-x-y\)
\(=x\left(2x-1\right)+y\left(2x-1\right)\)
\(=\left(x+y\right)\left(2x-1\right)\)
d. \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
e. \(x^2-2xy-x^2-4y^2\)
\(=-2xy-4y^2\)
\(=-2y\left(x+2y\right)\)
g. \(x^2-2x+1-16\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-1-4\right)\left(x-1+4\right)\)
Bài 2:
a, \(x^2-6x+10=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\ge1>0\)
\(\Rightarrowđpcm\)
b, \(x^2-4xy+4y^2+1=\left(x-2y\right)^2+1>0\)
\(\Rightarrowđpcm\)
c, \(x^2-4x+7=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\)
\(\Rightarrowđpcm\)
d, \(x^2+y^2-2x+4y+5\)
\(=x^2-2x+1+y^2+4y+4\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
\(\Rightarrowđpcm\)
Ép người quá đáng >.<
Bài 1:
a, \(-\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)+\left(2x^2+1\right)\)
\(=-\left(4x^4-4x^3+2x^2+4x^3-4x^2+2x+2x^2-2x+1\right)+2x^2+1\)
\(=-\left(4x^4+1\right)+2x^2+1=-4x^4+2x^2\)
b, \(\left(x^2+x+2\right)^2+\left(x-1\right)^2-2\left(x^2+x+2\right)\left(x-1\right)\)
\(=\left(x^2+x+2-x+1\right)^2=\left(x^2+3\right)^2\)
d, \(-125x^3+225x^2-135x+27\)
\(=-\left(125x^3-225x^2+135x-27\right)\)
\(=-\left(125x^3-75x^2-150x^2+90x+45x-27\right)\)
\(=-\left[25x^2\left(5x-3\right)-30x\left(5x-3\right)+9\left(5x-3\right)\right]\)
\(=-\left[\left(5x-3\right)\left(25x^2-15x-15x+9\right)\right]\)
\(=-\left(5x-3\right)^3\)
a)\(x^2+4y^2-2x+4y+2\)
\(=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\ge0\)(đúng)
b) Sửa đề
\(3y^2+x^2+2xy+2x+6y+3\)
\(=\left(x^2+y^2+2xy\right)+2y^2+2x+6y+3\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+2y^2+4y+2\)
\(=\left(x+y+1\right)^2+2\left(y+1\right)^2\ge0\) (đúng)