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a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Ta có :
\(x^6+3x^5-2x^4+7x^3-2x^2+3x+1\)
\(=x^6-x^5+x^4+4x^5-4x^4+4x^3+x^4-x^3+x^2+4x^3-4x^2+4x+x^2-x+1\)
\(=x^4\left(x^2-x+1\right)+4x^3\left(x^2-x+1\right)+x^2\left(x^2-x+1\right)+4x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^4+4x^3+x^2+4x+1\right)\)
a) 7x+7y=7(x+y)
b) 2x2y-6xy2=2xy(x-3y)
c)3x(x-1)+7x2(x-1)=x(x-1)(3+7x)
d)3x(x-4)+5x2(4-x)=(x-4)(3x-5x2)
=x(x-4)(3-5x)
e)6x4-9x3=3x3(2x-3)
f)5y8-15y6=5y6(y2-3)
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
a) x3+y3+z3-3xyz
=(x+y)3+z3-3x2y-3xy2-3xyz
=(x+y+z).[(x+y)2+(x+y).z+z2]-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2)-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2-3xy)
=(x+y+z)(x2+y2+zx+zy+z2-zy)
b)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(-a2c+c2a)+(b2c-b2a)
=b.(a2-c2)-ac.(a-c)-b2.(a-c)
=b.(a+c)(a-c)-ac.(a-c)-b2.(a-c)
=(a-c)[b.(a+c)-ac-b2]
=(a-c)(ab+bc-ac-b2)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
1) (3x+4)(x+1) = 3x2+7x+4 đặt là a
(6x+7)2= 36x2+84x+49 = 12a+1
=> a(12a+1)- 6 = 12a2 -a -6 = (3a+2)(4a-3) = (9x2+21x+14)(12x2+28x+13)
2) (x-2)2=x2-4x+4 đặt là a
(2x-5)(2x-3)= 4x2-16x+15 =4a-1
=> a(4a-1)-5 = 4a2-a-5 = (4a-5)(a+1) = ( 4x2-16x+11)(x2-4x+5)
3) đặt (x+3)2 =a ta làm tương tự
4) (x-2)(x-10)(x-4)(x-5) = (x2-12x+20)(x2-9x+20)
đặt x2+20=a => (a-12x)(a-9x)-54x2 = a2-21ax+54x2 = (a-18x)(a-3x) = (x2-18x+20)(x2-3x+20)
\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)
\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)
\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)
\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)