phan tich da thuc thanh nhan tu : x4+x2+1
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\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
4x^4 - 32x^2 +1 = 4x^4 + 4x^2 +1 - 36x^2 = (2x^2 + 1)^2 - 36x^2 = (2x^2 - 6x + 1)(2x^2 + 6x + 1)
4 x4 - 32 x2 + 1
= ( 2 x2 )2 - 2 . 2x2. 8 + 64 - 63
= ( 2 x2 - 8 )2 - 63
= ( 2x2 - 8 + √63 ) ( 2x2 - 8 - √63 )
Xong
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
Ta có : x5 + x + 1
= x5 + x4 - x4 - x3 + x3 + x2 - x2 - x + x + 1
= (x5 + x4) - (x4 + x3) + (x3 + x2) - (x2 + x) + (x + 1)
= x5(x + 1) - x4.(x + 1) + x3(x + 1) - x2(x + 1) + (x + 1)
= (x + 1)(x5 - x4 + x3 - x2 + 1)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
x^4 + x^2 + 1
= x^4 + 2x^2 + 1 - x^2
= ( x^2 + 1)^2 - x^2
= ( x^2 - x + 1 )( x^2 + x + 1)