K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 10 2016

a) \(x^4-2x^3+2x-1\)

\(=x^4-x^3-x^3+2x-2+1\)

\(=\left(x^4-x^3\right)+\left(2x-2\right)-\left(x^3-1\right)\)

\(=x^3\left(x-1\right)+2\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x-1\right)\left(x^3+2-x^2-x-1\right)\)

\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)

\(=\left(x-1\right)\left[\left(x^3-x^2\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left(x^2-1\right)\left(x-1\right)\)

\(=\left(x-1\right)^2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^3\left(x+1\right)\)

b) \(x^4+2x^3+2x^2+2x+1\)

\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)

\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+1+2x\right)\)

\(=\left(x^2+1\right)\left(x+1\right)^2\)

4 tháng 8 2017

a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)

=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)

=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)

b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)

=\(a^2b^2-2ab+1+a^2+2ab+b^2\)

=\(a^2b^2+a^2+b^2+1\)

=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)

=\(\left(b^2+1\right)\left(a^2+1\right)\)

c,\(x^4+2x^3+2x^2+2x+1\)

=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)

=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)

=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)

=\(\left(x+1\right)^2\left(x^2+1\right)\)

8 tháng 8 2018

\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)

\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)

\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)

\(=\left(x^2+1\right)\left(x+1\right)^2\)

\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)

8 tháng 8 2018

\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)

                                    \(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)

                                   \(=\left(x+1\right).\left(x^2+x+1\right)\)

\(x^3-4x^2+12x-27\)

\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)

\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)

\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)

\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)

\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)

19 tháng 8 2017

a)45+x3-5x2-9x

(45-9x)+(x3-5x2)

9(5-x)+x2(x-5)

(9-x2)(5-x)

(3-x)(3+x)(5-x)

b)x4-2x3-2x2-2x-3

x3(x-2)-2x(x-2)-3

(x-2)(x3-2x)-3

x

29 tháng 1 2019

\(x^4+2x^3+3x^2+2x+1.\)

\(=x^4+x^3+x^3+x^2+x^2+x^2+x+x+1\)
\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)

\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x+1\right)^2+x\left(x+1\right)^2+\left(x+1\right)^2\)

\(=\left(x+1\right)^2\left(x^2+x+1\right)\)

\(=\left(x+1\right)^2\left(x+1\right)^2\)

\(=\left(x+1\right)^4\)

29 tháng 1 2019

@wi

\(x^2+x+1=\left(x+1\right)^2???\)

\(x^2+2x+1=\left(x+1\right)^2\)chứ

3 tháng 9 2018

\(x^2-2x-4y^2-4y\)

\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

1 tháng 10 2020

\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}

26 tháng 9 2018

      \(x^3-x^2-14x+24\)

\(=x^3-2x^2+x^2-2x-12x+24\)

\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+x-12\right)\)

\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)

\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)

\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)

      \(x^4+x^3+2x-4\)

\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)

\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)

\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)

      \(8x^4-2x^3-3x^2-2x-1\)

\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)

\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)

\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)

\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)

\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)

\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)

      \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

Chúc bạn học tốt.