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b) \(64x^3+1=\left(4x+1\right)\left(16x^2-4x+1\right)\)\
c) \(x^3y^6z^9-125=\left(xy^2z^3-5\right)\left(x^2y^4z^6+5xy^2z+25\right)\)
d) \(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)
e) \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
64x3 + 1
= ( 4x )3 + 1
= ( 4x + 1 ) ( 16x2 - 4x + 1 )
Hằng đẳng thức 6 : A3 + B3
27x6 - 8x3
= ( 3x2)3 + ( 2x )3
= ( 3x + 2x ) ( 9x2 - 6x + 4x2 )
HĐT 6
x6 - y6
= ( x2 )3 - ( y2 )3
= ( x2 - y2 ) ( x4 + x2y2 + y4 )
HĐT 7 : A3 - B3
x3y6z9 + 1
= ( xy2z3)3 + 1
= ( xy2z3 + 1 ) ( x2y4z6 + zy2z3 + 1 )
HĐT 6
b: \(=\left(x^2+4x-3\right)^2-2x\left(x^2+4x-3\right)-3x\left(x^2+4x-3\right)+6x^2\)
\(=\left(x^2+4x-3\right)\left(x^2+4x-3-2x\right)-3x\left(x^2+4x-3-2x\right)\)
\(=\left(x^2+2x-3\right)\left(x^2+4x-3-3x\right)\)
\(=\left(x^2+x-3\right)\left(x+3\right)\left(x-1\right)\)
c: \(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-3b^2c+3bc^2-c^3+c^3-3a^2c+3ac^2-a^3\)
\(=-3a^2b+3ab^2-3b^2c+3bc^2-3a^2c+3ac^2\)
\(=-3\left(a^2b-ab^2+b^2c-bc^2+a^2c-ac^2\right)\)
\(\left(6x-1\right)^2-\left(3x+2\right)\)
\(=36x^2-12x+1-3x-2\)
\(=36x^2-15x-1\)
bn ktra lại đề nhé
hk tốt
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
b)3x^2-18x+27=3x^2-9x-9x+27=3x*(x-3)-9*(x-3)=(x-3)*(3x-9)=(x-3)*3*(x-3)=3*(x-3)^2
c)x^3-4x^2-12x+27=(x+3)*(x^2-3x+9-4)=(x+3)*(x^2-3x+5)
d)27x^3-1/27=(3x-1/3)*(9x^2-x+1/9) (hang dt)
con a) voi e) mk chiu
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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)\(x^2+4x-4y^2-8y\)
\(=x^2+2xy+4x-2xy-4y^2-8y\)
\(=x\left(x+2y+4\right)-2y\left(x+2y+4\right)\)
\(=\left(x-2y\right)\left(x+2y+4\right)\)
b)sai đề
c)sai đề tiếp
a)x2+4x-4y2-8y=(x2-4y2)+(4x-8y)
=(x+2y(x-2y)+4(x-2y)
=(x-2y)(x+2y+4)