Phân tích các đa thức sau thành nhân tử:
a) x3 - 2x2 + x b) x2 – 2x – 15
c) 5x2y3 – 25x3y4 + 10x3y3 d) 12x2y – 18xy2 – 30y2
e) 5(x-y) – y.( x – y) g)36 – 12x + x2
h) 4x2 + 12x + 9 i) 11x + 11y – x2 – xy
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a) x2 - 5x - y2 -5y
= ( x2 - y2 ) + ( -5x - 5y)
= ( x - y ) ( x + y) - 5( x + y )
= ( x + y ) ( x - y -5)
b) x3 + 2x2 - 4x - 8
= x2 ( x + 2 ) - 4 ( x + 2 )
= ( x +2 ) ( x2 -4 )
= ( x+2)2 ( x-2)
Bai 2 :
a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)
\(=2x^2+2x+13-2x^2-2x+12=25\)
b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)
\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)
\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)
\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)
\(M=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}\)ĐKXĐ : a khác 0, a khác 1
\(M=\frac{1+a}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}\)
\(M=\frac{a-1}{a}\)
\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)DK : \(x\ne0;\pm1\)
\(=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}=\left(\frac{1}{a\left(a-1\right)}+\frac{a}{a\left(a-1\right)}\right):\frac{a+1}{\left(a-1\right)^2}\)
\(=\frac{a+1}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}=\frac{a-1}{a}\)
x2 + 5y2 - 4xy + 6x - 14y + 10 = 0
=> (x2 - 4xy + 4y2) + (6x - 12y) + 9 + (y2 - 2y + 1) = 0
=> (x - 2y)2 + 6(x - 2y) + 9 + (y - 1)2 = 0
=> (x - 2y + 3)2 + (y - 1)2 = 0
=> \(\hept{\begin{cases}x-2y+3=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Vậy x = 1 ; y = - 1 là giá trị cần tìm
a) x3 - 2x2 + x
= x(x2 - 2x + 1)
= x(x - 1)2
b) x2 - 2x - 15
= x2 - 2x + 1 - 16
= (x - 1)2 - 42
= (x - 5)(x + 3)
c) 5x2y3 - 25x3y4 + 10x3y3
= 5x2y3(1 - 5xy + 2x)
d) 12x2y - 18xy2 - 30y2
= 6y(2x2 - 3xy - 5y)
= 6y(2x2 + 2xy - 5xy - 5y)
= 6y[2x(x + y) - 5y(x + y)
= 6y(x + y)(2x - 5y)
e) 5(x - y) - y(x - y)
= (5 - y)(x - y)
g) 36 - 12x + x2
= (6 - x)2
h) 4x2 + 12x + 9
= (2x + 3)2
i) 11x + 11y - x2 - xy
= 11(x + y) - x(x + y)
= (!1 - x)(x + y)
a, \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
b, \(x^2-2x-15=\left(x^2-2x+1\right)-16=\left(x-1\right)^2-4^2=\left(x-5\right)\left(x+3\right)\)
c, \(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
d, \(12x^2y-18xy^2-30y^2=3y\left(4x^2-6xy-10y\right)\)
\(=3y\left[2y\left(2y-3x-5\right)\right]=6y^2\left(2y-3x-5\right)\)
e, \(5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)
g, \(36-12x+x^2=\left(6-x\right)^2\)
h, \(4x^2+12x+9=\left(2x+3\right)^2\)
i, \(11x+11y-x^2-xy=11\left(x+y\right)-x\left(x+y\right)=\left(11-x\right)\left(x+y\right)\)