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\(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(-5x+1\right).\left(x-3\right)\)
\(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=\left(2x^2+2x\right)+\left(5x+5\right)\)
\(=2x.\left(x+1\right)+5.\left(x+1\right)\)
\(=\left(2x+5\right).\left(x+1\right)\)
\(2x^2+3x+5\) (Bạn xem lại đề nhé.)
\(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right).\left(x^2-x+1\right)-3x.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right).\left(x^2-4x+1\right)\)
\(x^2-4x-5\)
\(=x^2-5x+x-5\)
\(=\left(x^2-5x\right)+\left(x-5\right)\)
\(=x.\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+1\right).\left(x-5\right)\)
\(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2-2a+1\right).\left(a^2+2a+1\right)\)
\(=\left(a-1\right)^2.\left(a+1\right)^2\)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
Hãy tích cho tui đi
khi bạn tích tui
tui không tích lại bạn đâu
THANKS
a, 3x3-8x2+8x-5
= x2(3x-5)-x(3x-5)+3x-5
=(3x-5)(x2-x+1)
b, 4x3-3x2+5x-21
= x2(4x-7) +x(4x-7)+3(4x-7)
=(4x-7)(x2+x+3)
a, \(2x^2+2x+5x+5=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)
b,\(2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
c,\(x^3-3x^2+1-3x=\left(x^3+1\right)-3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d,\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
e,\(\left(a^2+1\right)^2-4a^2=\left(a^2+1\right)^2-\left(2a\right)^2=\left(a^2-2a+1\right)\left(a^2+2a+1\right)=\left(a-1\right)^2\left(a+1\right)^2\)