\(\text{Phân tích đa thức thành nhân tử (bằng kĩ thuật tách hạng tử)}\):\( -8x^2 +23x + 3\)
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a) -8x2+5x+3=-8x^2+8x-3x+3=-8x(x-1)-3(x-1)=-(8x+3)(x-1)
b)8x^2-10x-3=8x^2-12x+2x-3=8x(x-1,5)+2(x-1,5)=2(4x+1)(x-1,5)
c)=8x^2-2x+12x-3=2x(4x-1)+3(4x-1)=(2x+3)(4x-1)
d)=-8x^2+24x-x+3=-8x(x-3)-(x-3)=-(8x+1)(x-3)
Bài làm:
Ta có: \(3x^2+3x-6\)
\(=\left(3x^2+6x\right)-\left(3x+6\right)\)
\(=3x\left(x+2\right)-3\left(x+2\right)\)
\(=3\left(x-1\right)\left(x+2\right)\)
\(3x^2+3x-6\)
\(=3\left(x^2+x-2\right)\)
\(=3\left(x^2+2x-x-2\right)\)
\(=3\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=3\left(x-1\right)\left(x+2\right)\)
\(1,2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
\(2,4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(\left(4x+1\right)\left(x-2\right)\)
3x^2 - 8x + 4
= 3x^2 - 6x - 2x + 4
=( 3x^2 - 6x ) - ( 2x - 4)
=3x(x-2) - 2(x-2)
=(3x-2) - (x-2)
a, = (x^3-x^2)-(4x^2-4x)+(4x-4)
= (x-1).(x^2-4x+4) = (x-1).(x-2)^2
b, = (x^3+x^2)-(10x^2+10x)+(16x+16)
= (x+1).(x^2-10x+16)
= (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)
k mk nha
a)= (x^3-x^2)-(4x^2-4x)+(4x-4)
= (x-1).(x^2-4x+4)
= (x-1).(x-2)^2
b)= (x^3+x^2)-(10x^2+10x)+(16x+16)
= (x+1).(x^2-10x+16)
= (x+1).[ (x^2-2x)-(8x-16) ]
= (x+1).(x-2).(x-8)
P/s tham khảo nha
a. \(=6x^2-4x-9x+6=2x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(2x-3\right)\)
b. \(=6x^2+4x+9x+6=2x\left(3x+2\right)+3\left(3x+2\right)=\left(3x+2\right)\left(2x+3\right)\)
a) 8x2 - 2x - 1
=8x2+2x-4x-1
=2x.(4x+1)-(4x+1)
=(4x+1)(2x-1)
b) x2 - y2 + 10x - 6y + 16
=x2+10x+25-y2-6y-9
=(x+5)2-(y+3)2
=(x+5-y-3)(x+5+y+3)
=(x-y+2)(x+y+8)
Ta có: \(-8x^2+23x+3\)
\(=\left(-8x^2+24x\right)-\left(x-3\right)\)
\(=-8x\left(x-3\right)-\left(x-3\right)\)
\(=\left(-8x-1\right)\left(x-3\right)\)
\(=\left(3-x\right)\left(8x+1\right)\)
\(-8x^2+23x+3\)
\(=-\left(8x^2-23x-3\right)\)
\(=-\left(8x^2-24x+x-3\right)\)
\(=-\left[8x\left(x-3\right)+\left(x-3\right)\right]\)
\(=-\left(8x+1\right)\left(x-3\right)\)