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\(x^4-x+2008x^2+2008x+2008\)
\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
x^2 + 7x -15
= x^2 + 7x +12,25 -27,25
= (x+3,5)^2 - 27, 25
= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))
\(B=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=\left(x^2+3x-1\right)^2\)
x3 - 7x - 6 = x3 - 3x2 + 3x2 - 9x + 2x - 6
= x2 (x - 3) + 3x(x - 3) +2(x - 3)
= (x2 + 3x + 2)(x - 2)
= (x2 + 2x +x + 2)(x - 2)
= (x + 2)(x + 1)(x - 2)
a)\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=\left(x^2-2x\right)+\left(3x-6\right)\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
a) x2 + x - 6
= x2 - 2x + 3x - 6
= (x2 - 2x) + (3x - 6)
= x(x - 2) + 3(x - 2)
= (x + 3)(x - 2)
b) x4 + 4
= x4 + 4x2 + 4 - 4x2
= (x4 + 4x2 + 4) - 4x2
= (x + 2)2 - 4x2
= (x + 2 - 2x)(x + 2 +2x)
b ( x^2 + 3x + 2)( x^2 + 7x + 12) - 24
= [ x^2 +x + 2x + 2) ( x^2 +3x + 4x + 12) - 24
= [x(x+1) + 2 (x + 1) [x(x+3) + 4(x+3) ] - 24
= ( x + 1)(x+2) (x+3)(x+4) - 24
= ( x + 1).(x+4) (x+2)(x+3) - 24
=(x^2 + 5x + 4)(x^2+5x+6) - 24
Đặt x^2 + 5x +4 =y ta có:
= y(y+2) - 24
= y^2 + 2y - 24
= y^2 + 2y + 1 - 25
= ( y + 1)^2 - (5)^2
= ( y + 1 - 5 )( y + 1 + 5)
= ( y- 4)(y +6)
Thay y trở lại là đc
đúng nha
a, x^2 + 7x + 6
= x^2 + x + 6x + 6
= x(x + 1) + 6(x + 1)
= (x + 6)(x + 1)
\(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+6\right)\left(x+1\right)\)