Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
a: \(50x^5-8x^3\)
\(=2x^3\left(25x^2-4\right)\)
\(=2x^3\left(5x-2\right)\left(5x+2\right)\)
b: \(x^4-5x^2-4y^2+10y\)
\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)
\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)
c: \(36a^2+12a+1-b^2\)
\(=\left(6a+1\right)^2-b^2\)
\(=\left(6a+1-b\right)\left(6a+1+b\right)\)
d: \(x^3+y^3-xy^2-x^2y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\cdot\left(x-y\right)^2\)
e: Ta có: \(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
f: Ta có: \(9x^4+16x^2-4\)
\(=9x^4+18x^2-2x^2-4\)
\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(9x^2-2\right)\)
g: Ta có: \(-6x^2+5xy+4y^2\)
\(=-6x^2+8xy-3xy+4y^2\)
\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left(-2x-y\right)\)
h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)
\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)
\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)
\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
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: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)
\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-3x+2y\right)\)
\(=0\cdot0\)
\(=0\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x+y\right)\left(x-y\right)\)
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
\(a,16x-5x^2-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
\(b,x^2-4x-5\)
\(=x^2+x-5x-5\)
\(=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
x2 - 4x - 5
= x2 - x + 5x - 5
= x ( x - 1 ) + 5 ( x - 1 )
= ( x - 1 ) ( x + 5 )