x3 + y3 + x2 – 2xy + 2y2
phân tích các đa thức thành nhân tử
a) x2-2xy +y2-z2
b) x3+y3+2x2-2xy+2y2
\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)
\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)
Phân tích đa thức thành nhân tử:
a) 2y2-3y-5
b) x2-9x-10
c) x3+y3+z3-3xyz
\(a,=\left(2x-5\right)\left(x+1\right)\\ b,=\left(x-10\right)\left(x+1\right)\\ c,=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
1) x3-x2+2x-2 4) ax-2x-a2+2a 7) x2-6xy-25z2+9y2
2) x2-y2+2x+2y 5) 2xy +3z+6y+xz 8) x3-2x2+x
3) x2/4+2xy+4y2-25 6) x2y2+yz+y3+zx2 9) x4+4
c) 3x + 3y – x2 – 2xy – y2 d) x3 – x + 3x2y + 3xy2 – y + y3
c) \(3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)d) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
\(c,=3\left(x+y\right)-\left(x+y\right)^2=\left(3-x-y\right)\left(x+y\right)\\ d,=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
d) \(\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Tính giá trị biểu thức:
a) M=x2-2xy+y2-10x+10y với x-y=9
b) N=x3+3x2y+3xy2+y3+x2+2xy+y2 với x=10-y
a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)
\(=\left(x-y\right)^2-10\left(x-y\right)\)
\(=9^2-10\cdot9=-9\)
Bài 1: Khai triển các hằng đẳng thức.
1,(x+1)2
2,(2x+1)2
3, (2x+y)2
4, (2x+3)2
5, ( 3x+2y)2
6, (2x2+1)2
7, (x3+1)2
8, (x2+y3)2
9, ( x2+2y2)2
10, (1/2x+1/3y)2
1) \(\left(x+1\right)^2=x^2+2x+1\)
2) \(\left(2x+1\right)^2=4x^2+4x+1\)
3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)
4) \(\left(2x+3\right)^2=4x^2+12x+9\)
5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)
6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)
7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)
8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)
9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)
10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)
Biết hệ phương trình x 3 + y 3 = 8 x + y + 2 x y = 2 có hai nghiệm ( x 1 ; y 1 ) ; ( x 2 ; y 2 ) . Tổng x 1 + x 2 bằng?
A. 2
B. −2
C. 1
D. 0
Bài 3: Phân tích đa thức sau thành nhân tử.
a) x4 + 2x2 + 1
b) 4x2 - 12xy + 9y2
c) -x2 - 2xy - y2
d) (x + y)2 - 2(x + y) + 1
e) x3 - 3x2 + 3x - 1
g) x3 + 6x2 + 12x + 8
h) x3 + 1 - x2 - x
k) (x + y)3 - x3 - y3
a) x⁴ + 2x² + 1
= (x²)² + 2.x².1 + 1²
= (x² + 1)²
b) 4x² - 12xy + 9y²
= (2x)² - 2.2x.3y + (3y)²
= (2x - 3y)²
c) -x² - 2xy - y²
= -(x² + 2xy + y²)
= -(x + y)²
d) (x + y)² - 2(x + y) + 1
= (x + y)² - 2.(x + y).1 + 1²
= (x - y + 1)²
e) x³ - 3x² + 3x - 1
= x³ - 3.x².1 + 3.x.1² - 1³
= (x - 1)³
g) x³ + 6x² + 12x + 8
= x³ + 3.x².2 + 3.x.2² + 2³
= (x + 2)³
h) x³ + 1 - x² - x
= (x³ + 1) - (x² + x)
= (x + 1)(x² - x + 1) - x(x + 1)
= (x + 1)(x² - x + 1 - x)
= (x + 1)(x² - 2x + 1)
= (x + 1)(x - 1)²
k) (x + y)³ - x³ - y³
= (x + y)³ - (x³ + y³)
= (x + y)³ - (x + y)(x² - xy + y²)
= (x + y)[(x + y)² - x² + xy - y²]
= (x + y)(x² + 2xy + y² - x² + xy - y²)
= (x + y).3xy
= 3xy(x + y)
a) Cho x+y=9,xy=18 tính x3+y3, x4+y4,x3-y3
b)Cho x+y = -9 ,tính A= x2+2xy+y2-6x-5y-5
Lời giải:
a.
$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=[9^2-2.18]^2-2.18^2=1377$
Nếu $x\geq y$ thì:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$
$=\sqrt{9^2-4.18}(9^2-18)=189$
Nếu $x< y$ thì $x^3-y^3=-189$
b.
$A=(x+y)^2-6(x+y)+y-5$
$=(-9)^2-6(-9)+y-5=130+y$
Chưa đủ cơ sở để tính biểu thức.
a) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=9^3-3\cdot18\cdot9=243\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left[\left(x+y\right)^2-2xy\right]^2-2\left(xy\right)^2\)
\(=\left(9^2-2\cdot18\right)^2-2\cdot18^2\)
\(=45^2-2\cdot324\)
=1377