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b) x3y3 + x2y2+ 4 = x3y3- 4xy + (xy)2- 2xy.2 + 22 = xy [ (xy)^2 - 2^2 ] + ( xy - 2)^2
= xy(xy-2)(xy+2)+ (xy-2)^2
= (xy-2) [ xy(xy+2) + ( xy-2) ]
= (xy-2) [ (xy)2 + 2xy + xy - 3 ]
= ( xy - 3) [ (xy)2 + 3xy - 3]
3) (chưa bik làm)
4) x4 +x3 + 6x2 +5x +5
= x4 +x3 + x2 + 5x2 + 5x +5
= x2( x2+x+ 1 ) + 5( x2+x+ 1 )
= ( x2+ 5 ) ( x2+x+ 1 )
5) x4 - 2x3 - 12x2 +12x + 36
= x4 - 2x3 - 6x2 - 6x2 + 12x + 36=
x2 ( x2 - 2x - 6) - 6 ( x2 - 2x - 6)
= (x^2 - 6) ( x2 - 2x - 6) 6) x8y8 + x4y4 + 1 = \(\left[\left(xy\right)^4\right]^2+2x^4y^4+1-x^4y^4\)=\(\left[\left(xy\right)^4+1\right]^2-\left[\left(xy\right)^2\right]^2\)
= \(\left(x^4y^4+1-x^2y^2\right)\left(x^4y^4+1+x^2y^2\right)\)
( mik ko bik đúng hay sai đâu nha) mik thấy nó thành nhân tử thì mik tách thôi
\(\text{a) }x^3y^3+x^2y^2+4\)
\(=x^3y^3+2x^2y^2-x^2y^2+4\)
\(=\left(x^3y^3+2x^2y^2\right)-\left(x^2y^2-4\right)\)
\(=x^2y^2\left(xy+2\right)-\left(xy+2\right)\left(xy-2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
\( {c)}\)\(x^4+x^3+6x^2+5x+5\)
\(=\left(x^4+x^3+x^2\right)+\left(5x^2+5x+5\right)\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+5\right)\)
\({d)}\)\(x^4-2x^3-12x^2+12x+36\)
\(=\left(x^4-2x^3-6x^2\right)-\left(6x^2-12x-36\right)\)
\(=x^2\left(x^2-2x-6\right)-6\left(x^2-2x-6\right)\)
\(=\left(x^2-2x-6\right)\left(x^2-6\right)\)
Câu b sai đề thì phải ah
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
x^3 - 7x - 6 = x^3 + x^2 - x^2 - 6x - x - 6 = (x^3 + x^2) - (x^2 + x) - (6x + 6)
= x^2(x + 1) - x(x + 1) - 6(x + 1) = (x + 1)(x^2 - x - 6) = (x + 1)(x^2 - 3x + 2x - 6)
= (x + 1){(x^2 - 3x) + (2x - 6)} = (x + 1){(x(x - 3) + 2(x - 3)}
= (x + 1)(x - 3)(x + 2)
\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
a)\(x^3y^3+x^2y^2+4\)
\(=x^3y^3-x^2y^2+2xy+2x^2y^2-2xy+4\)
\(=xy\left(x^2y^2-xy+2\right)+2\left(x^2y^2-xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
b)\(x^4+x^3+6x^2+5x+5\)
\(=x^4+x^2+x^2+5x^2+5x+5\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)
\(=\left(x^2+5\right)\left(x^2+x+1\right)\)
c)\(x^4-2x^3-12x^2+12x+36\)
\(=x^4-2x^3-6x^2-6x^2+12x+36\)
\(=x^2\left(x^2-2x-6\right)-6\left(x^2-2x-6\right)\)
\(=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d)\(x^8y^8+x^4y^4+1\)
\(=x^8y^8+2x^4y^4+1-x^4y^4\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1+x^2y^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(x^4y^4+2x^2y^2+1-x^2y^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(\left(x^2y^2+1\right)^2-\left(xy\right)^2\right)\left(x^4y^4+1-x^2y^2\right)\)
\(=\left(x^2y^2+1-xy\right)\left(x^2y^2+1+xy\right)\left(x^4y^4+1-x^2y^2\right)\)
bạn lm pb = cách nhẩm nghiệm đc không