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Bạn hội con bò gì đó ơi cho mk tham gia đc không vì là hội học hành nên .....
A) 1/2 x(x^2-4)+4(x+2)
=1/2x(x-2)(x+2)+4(x+2)
=(x+2)(1/2x^2-x+4)
b) 21(x-y)^2-7(x-y)^3
= (x-y)^2(21-7x+7y)
=(x-y)^2.7(3-x+y)
c) 1/8x^3-3/4x^2+3/2x-1
=(1/2x)^3-3.(1/2x)^2.1+3.1/2x.1^2-1
=(1/2x-1)^3
a)\(2a^3+16=2\left(a^3+8\right)=2\left(a+2\right)\left(a^2-2a+4\right)\)
b)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+\left(3y\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2\)
\(=\left(2x+3y\right)^2\)
c)\(x^4-2x^3-x^2+2x+1=\left(x^4-x^3-x^2\right)-\left(x^3-x^2-x\right)-\left(x^2-x-1\right)\)
\(=x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)-\left(x^2-x-1\right)\)
\(=\left(x^2-x-1\right)\left(x^2-x-1\right)=\left(x^2-x-1\right)^2\)
\(\left(x-5\right)^2-16=\left(x-5\right)^2-4^2=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)
\(25-\left(3-x\right)^2=5^2-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
\(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)\(49\left(y-4\right)^2-9\left(y+2\right)^2=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left(7y-28-3y-6\right)\left(7y-28+3y-6\right)=\left(4y-34\right)\left(10y-22\right)\)\(=4.\left(2y-17\right)\left(5y-11\right)\)
e); f) Áp dụng hằng đẳng thức số 6,7 để làm
\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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) x4 - x3y - x + y
= x3(x - y) - (x - y)
=(x - y)(x3 - 1)
=(x - y)(x - 1)(x2 + x + 1)
b) x3 - 4x2 - 8x + 8
= (x3 + 8) - (4x2 + 8x)
= (x + 2)(x2 - 2x + 4) - 4x(x + 2)
= (x + 2)(x2 - 2x + 4 - 4x)
= (x + 2)(x2 - 6x + 4)
a, = (x^4-x^3y)-(x-y)
=x^3.(x-y)-(x-y) = (x-y).(x^3-1) = (x-y).(x-1).(x^2+x+1)
b, = (x^3+2x^2)-(6x^2+12x)+(4x+8)
= x^2.(x+2)-6x.(x+2)+4.(x+2) = (x+2).(x^2-6x+4)
k mk nha