phân tích đa thức thành nhân tử
\(A=x^8-2x^4-8\)
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a: \(x^4-2x^3+x^2-2x\)
\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+1\right)\)
b: \(x^4+x^3-8x-8\)
\(=\left(x^4+x^3\right)-\left(8x+8\right)\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-8\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
x^3+x^2-2x-8
= (x-2)(x^2+3x+4)
nah bạn chúc bạn học tốt nha
x3 + x2 - 2x - 8
= ( x3 - 8 ) + ( x2 - 2x )
= ( x - 2 ) . ( x2 + 2x + 4 ) + x ( x - 2 )
= ( x - 2 ) .( x2 + 2x + 4 + x )
= ( x-2 ) . ( x2 + 3x + 4 )
\(x^3-8+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4+2x\right)=\left(x-2\right)\left(x^2+4x+4\right)\\ =\left(x-2\right)\left(x+2\right)^2\)
=\(\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)
=\(\left(x-2\right)\left(x^2+4x+4\right)\)
=\(\left(x-2\right)\left(x+2\right)^2\)
a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)
b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)
Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)
c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)
Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)
d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)
Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)
\(A=x^8-2x^4-8\)
\(A=x^8-2x^4+4x^4-8\)
\(A=x^4\left(x^4-2\right)-4\left(x^4-2\right)\)
\(A=\left(x^4-4\right)\left(x^4-2\right)\)
\(a=\left(x^2-2\right)\left(x^2+2\right)\left(x^4-2\right)\)
\(A=\left(x^4\right)^2-4x^4+2x^4-8\)
\(=x^4\left(x^4-4\right)+2\left(x^4-4\right)\)
\(=\left(x^4+2\right)\left(x^4-4\right)\)
\(=\left(x^4+2\right)\left(x^2+2\right)\left(x^2-2\right)\)
\(=\left(x^4+2\right)\left(x^2+2\right)\left(x-2\right)\left(x+2\right)\)
Bạn cũng có thể đặt \(t=x^4\)để bài toán dễ làm hơn