Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a. $3x^3-12x^2+12x=3x(x^2-4x+4)=3x(x-2)^2$
b. $x^2-25+4xy+4y^2=(x^2+4xy+4y^2)-25=(x+2y)^2-5^2=(x+2y-5)(x+2y+5)$
c. $4x^3-x=x(4x^2-1)=x[(2x)^2-1^2]=x(2x-1)(2x+1)$
d. $x^2-x+2y-4y^2=(x^2-4y^2)-(x-2y)=(x-2y)(x+2y)-(x-2y)=(x-2y)(x+2y+1)$
Bài 2:
a. $3x(x-1)+x-1=0$
$\Leftrightarrow (x-1)(3x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $3x+1=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{-1}{3}$
b. $x(2x+1)-4x^2+1=0$
$\Leftrightarrow x(2x+1)-(4x^2-1)=0$
$\Leftrightarrow x(2x+1)-(2x-1)(2x+1)=0$
$\Leftrightarrow (2x+1)[x-(2x-1)]=0$
$\Leftrightarrow (2x+1)(-x+1)=0$
$\Leftrightarrow 2x+1=0$ hoặc $-x+1=0$
$\Leftrightarrow x=\frac{-1}{2}$ hoặc $x=1$
1/ x^2 +4xy +4y^2 = (x +2y)^2
2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3
3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3
4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3
Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
a: \(=2x^6-x^5-\dfrac{1}{3}x^4\)
b: \(=4xy^2-x^3+y^2-\dfrac{3}{4}x^2y\)
c: \(\left(3x^3-2xy^3+4y^2\right)\cdot\left(\dfrac{1}{6}x^2y^2\right)\)
\(=\dfrac{1}{2}x^5y^2-\dfrac{1}{3}x^3y^5+\dfrac{2}{3}x^2y^4\)
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
a) \(x^2+4x+4-y^2\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)