phân tích đa thức thành nhân tử (dùng hằng đẳng thức
bài 1
a,4x4-4x2y+y2 b,8x6-\( {1 \over 27}\)
c,x4-y4 d,1+6x2+12x4+8x6
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a) 6x² + 7xy + 2y²
= 6x² + 4xy + 3xy + 2y²
= (6x² + 4xy) + (3xy + 2y²)
= 2x(3x + 2y) + y(3x + 2y)
= (3x + 2y)(2x + y)
b) x² - y² + 10x - 6y + 16
= x² + 10x + 25 - y² - 6y - 9
= (x² + 10x + 25) - (y² + 6y + 9)
= (x + 5)² - (y + 3)²
= (x + 5 - y - 3)(x + 5 + y + 3)
= (x - y + 2)(x + y + 8)
c) 4x⁴ + y⁴
= 4x⁴ + 4x²y² + y⁴ - 4x²y²
= (2x² + y²)² - (2xy)²
= (2x² + y² - 2xy)(2x² + y² + 2xy)
a) \(=\left(x-2\right)^2\)
b) \(=\left(2x+1\right)^2\)
c) \(=\left(4x-3y\right)\left(4x+3y\right)\)
d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)
e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)
f) \(=\left(x-y\right)\left(x^2+xy+y^2\right)\)
g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)
h) \(=\left(x+2\right)^3\)
i) \(=\left(1-x\right)^3\)
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10x\right)\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
1.
$4x^2y+5x^3-x^2y^2=x^2(4y+5x-y^2)$
2.
$5x(x-1)-3y(1-x)=5x(x-1)+3y(x-1)=(x-1)(5x+3y)$
3.
$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5)$
4.
$6x-9-x^2=-(x^2-6x+9)=-(x-3)^2$
5.
$x^2+4y^2+4xy=x^2+2.x.2y+(2y)^2=(x+2y)^2$
6.
$\frac{1}{64}-27x^3=(\frac{1}{4})^3-(3x)^3$
$=(\frac{1}{4}-3x)(\frac{1}{16}+\frac{3x}{4}+9x^2)$
7.
$x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3$
$=(x-2)^3$
8.
$x^2-x-y^2-y=(x^2-y^2)-(x+y)=(x-y)(x+y)-(x+y)$
$=(x+y)(x-y-1)$
9.
$5x-5y+ax-ay=5(x-y)+a(x-y)$
$=(x-y)(5+a)$
Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)
\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)
CÁC BẠN ƠI CHỖ KIA LÀ 1/27 NHÉ MIK VIẾT NHẦM
a) 4x4 - 4x2y + y2 = ( 2x - y )2
b) 8x6 - 1/27 = ( 2x2 )3 - ( 1/3 )3 = ( 2x2 - 1/3 )( 4x4 + 2/3x2 + 1/9 )
c) x4 - y4 = ( x2 )2 - ( y2 )2 = ( x2 - y2 )( x2 + y2 ) = ( x + y )( x - y )( x2 + y2 )
d) 1 + 6x2 + 12x4 + 8x6 = ( 2x2 + 1 )3