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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!!!
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
a)\(81x^2-6yz-9y^2-z^2\)
\(=81x^2-\left(z-3y\right)^2\)
\(=\left(9x-z+3y\right)\left(9x+z-3y\right)\)
b)\(x^2y-x^3-9y+9x\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
c)\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4z^2\right)\)
\(=3\left[\left(a-b\right)^2-4z^2\right]\)
\(=3\left(a-b-2z\right)\left(a-b+2z\right)\)
a)\(81x^2-6yz-9y^2-z^2=\left(9x\right)^2-\left(9y^2+6yz+z^2\right)=\left(9x\right)^2-\left(3y+z\right)^2=\left(9x-3y-z\right)\left(9x+3y+z\right)\)b)\(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x^2-9\right)\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
c)\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
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)
=5(x2-\(\frac{18}{5}x\)-\(\frac{8}{5}\))=5(x2+\(\frac{2}{5}x-4x-\frac{8}{5}\))
=5(x-4)(\(\frac{2}{5}+x\))
= x^4 + 7x^3 - x^3 - 7x^2 - 11x^2 - 77x - 4x - 28
= x^3 ( x + 7 ) - x^2 ( x+ 7 ) - 11x( x+ 7 ) - 4 ( x+ 7 )
= ( x+ 7 )( x^3 -x^2- 11x - 4 )
Tự làm tiếp