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a) co sai de ko
b)x3-2x2+4x2-8x+3x-6=x2(x-2)+4x(x-2)+3(x-2)=(x-2)(x2+4x+3)=(x-2)(x+3)(x+1)
c)x3-2x2+2x2-4x-3x+6=x2(x-2)+2x(x-2)-3(x-2)=(x-2)(x2+2x-3)=(x-2)(x+3)(x-1)
d)x3-3x2+x2-3x-2x+6=x2(x-3)+x(x-3)-2(x-3)=(x-3)(x2+x-2)=(x-3)(x+2)(x-1)
a) \(x^2\left(x-3\right)+27-9x=0\)
\(x^2\left(x-3\right)+9\left(3-x\right)=0\)
\(x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\left(x^2-9\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=9\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=3\end{cases}}\Rightarrow x=3\)
vay \(x=3\)
a) \(x^2-4x+3\)
= \(x^2-3x-x+3\)
\(=\left(x^2-3x\right)-\left(x-3\right)\)
\(=x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
a)\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(x-1\right)\)
b)\(x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)
c)\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)
d)\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a/ \(x^3-5x^2+8x-4\)
= \(\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
b/ \(x^3-x^2+x-1\)
= \(\left(x^3-x^2\right)+\left(x-1\right)\)
= \(x^2\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+1\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)
\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)
\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)