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\(=\left(3x+15\right)^2-\left(x-7\right)^2=\left(4x+8\right)\left(2x+22\right)=8\left(x+2\right)\left(x+11\right)\)
(x-7)^2 = x^2-14x+49
<=> 9x^2+90x+225 -x^2+14x-49
= 8x^2+104x+176
= 8(x^2+13+22)
<=> 8(x+2)(x+11)
9x2+90x+225-(x-7)2
=(3x+15)2-(x-7)2
=(3x+15-x+7)(3x+15+x-7)
=(2x+22)(4x+8)
=2 (x+11)4 (x+2)
=8 (x+11)(x+2)
9x2 + 90x + 225 - ( x - 7 )2
= ( 3x + 15 )2 - ( x - 7 )2
= ( 3x + 15 + x - 7 )( 3x + 15 - x + 7 )
= ( 4x + 8 )( 2x + 22 )
= 4( x + 2 ) 2 ( x + 11 )
= 8( x + 2 )( x + 11 )
Hk tốt
a) bt \(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x+1\right)\left(x-2\right)\)
kl: ...
b) \(=\left(x+2\right)\left(x^2-8x-15\right)=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
kl:....
a, \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x^2-2x+x-2\right)\)
\(=\left(x-8\right)\left[x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b, \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left(x^2-3x+2x-6\right)\)
\(=\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
Chúc bạn học tốt!!!
\(x^3+\frac{1}{x^3}=x^3+\left(\frac{1}{x}\right)^3=\left(x+\frac{1}{x}\right)\left(x^2-x+\frac{1}{x^2}\right)\)( x khác 0 )
\(-x^3+9x^2-27x+27=-\left(x^3-9x^2+27x-27\right)=-\left(x-3\right)^3\)
\(\left(xy+1\right)^2-\left(x-y\right)^2=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
a) ( 4x+1) (12x-1) (3x+2) (x+1) -4
=(4x+1)(3x+2)(12x-1)(x+1)-4
=(12x2+11x+2)(12x2+11x-1)-4
Đặt t=12x2+11x+2 ta được:
t.(t-3)-4
=t2-3t-4
=t2+t-4t-4
=t.(t+1)-4.(t+1)
=(t+1)(t-4)
thay t=12x2+11x+2 ta được:
(12x2+11x+3)(12x2+11x-2)
Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x2+11x+3)(12x2+11x-2)
b) (x2+2x)2+9x2+18x+20
=(x2+2x)2+9.(x2+2x)+20
Đặt y=x2+2x ta được:
y2+9y+20
=y2+4y+5y+20
=y.(y+4)+5.(y+4)
=(y+4)(y+5)
thay y=x2+2x ta được:
(x2+2x+4)(x2+2x+5)
Vậy (x2+2x)2+9x2+18x+20=(x2+2x+4)(x2+2x+5)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2+4-4x\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)
9x2+90x+225-(x-7)2
=(3x)2+2.3x.15+152-(x-7)2
=(3x+15)2-(x-7)2
=[(3x+15)-(x-7)][(3x+15)+(x-7)]
=(3x+15-x+7)(3x+15+x-7)
=(2x+22)(4x+8)
=2(x+11).4(x+2)
=8(x+11)(x+2)