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\(a,2x\left(x-3\right)\\ b,x^2-\left(y+1\right)^2\\ =\left(x-y-1\right)\left(x+y+1\right)\)

21 tháng 10 2022

a,2x(x−3)

b,x^2−(y+1)T^2=(x−y−1)(x+y+1)

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

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

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

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

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

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

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

a)\(x^2+2x-3=x^2+3x-x-3\) 

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

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

23 tháng 12 2018

a) \(x^3-27+2x^2-6x\)

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

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

23 tháng 12 2018

b) \(18xy-12xy^2+2xy^3\)

\(=2xy\left(9-6y+y^2\right)\)

\(=2xy\left(y-3\right)^2\)

9 tháng 10 2016

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

=>\(x\left(6x^2+x-2\right)\)

19 tháng 10 2016

phần 1 đề nhầm ak sửu lại nha:

\(\left(8x^3+1\right):\left(4x^2-2x+1\right)=\left(2x+1\right)\left(4x^2-2x+1\right):\left(4x^2-2x+1\right)=2x+1\)

2) \(x^2-y^2-6x+6y\)

\(=\left(x-y\right)\left(x+y\right)-6\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-6\right)\)

19 tháng 10 2016

a) \(\left(8x^3+1\right):\left(4x^2-2x+1\right)\)

 

3 tháng 7 2018

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

2, \(=2\left(x^2-y^2\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1+4\right)=2\left(x+1\right)\left(x+3\right)\)

3, \(=\left(x^2+6x+9\right)-\left(2y\right)^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)

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

30 tháng 9 2018

\(4y^2-x^2+2x-1\)

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

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

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

hk tốt

^^

26 tháng 7 2018

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

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

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

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

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

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

30 tháng 9 2018

\(x^2-y^2+2x+1\)

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

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

\(=\left(x-y+1\right)\left(x+y+1\right)\)

hk tốt

^^

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}