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a) \(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2+1\right)\)
b) \(4x^2-4x-3=4x^2+2x-6x-3=2x\left(2x+1\right)-3\left(2x+1\right)=\left(2x+1\right)\left(2x+3\right)\)
c) \(4x^4+81=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x-9\right)\)
d) \(x^2-6xy-25+9y^2=\left(x-3y\right)^2-25=\left(x-3y-5\right)\left(x-3y+5\right)\)
e) \(x^2-8y^2-2xy=x^2+2xy-4xy-8y^2=x\left(x+2y\right)-4y\left(x+2y\right)=\left(x+2y\right)\left(x-4y\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
a) 3x2 - 7x + 2
= 3x2 - 6x - x + 2
= (3x2 - 6x) - (x - 2)
= 3x (x - 2) - (x - 2)
= (3x - 1) (x - 2)
a) (x^2+x)^2-14(x^2+x)+24
=(x^2+x)^2-2(x^2+x)-12(x^2+x)24
=(x^2+x)(x^2+x-2)-12(x^2+x-2)
=(x^2+x-12)(x^2+x-2)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a);b);c) Dùng máy tính (cụ thể là solve) bấm nghiệm rồi phân tích
d)Nhóm số T1;T2;T4 lại vs nhau
e)Biến đổi thành x2-2xy+y2-9y2