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\(B=x^2-2x+y^2+4y+6=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\forall x,y\)
\(B=x^2-2x+y^2+4y+6\)
\(=x^2-2x+1+y^2+4y+4+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\forall x,y\)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
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
\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
a) \(A=x^2-x+1=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
b) \(B=\left(x-2\right)\left(x-4\right)+3=x^2-6x+8+3=\left(x-3\right)^2+2\ge2>0\)
c) \(C=2x^2-4xy+4y^2+2x+5=\left(x-2y\right)^2+\left(x+1\right)^2+4\ge4>0\)
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)