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23 tháng 10 2016

kết quả thôi nha

23 tháng 10 2016

umk nhanh nha bạn

11 tháng 10 2020

a) \(4x^3y-12x^2y^3-8x^4y^3\)

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

b) \(2x^2+4x+2-2y^2\)

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

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

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

c) \(x^3-2x^2+x-xy^2\)

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

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

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

d) \(x\left(x-2y\right)+3\left(2y-x\right)\)

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

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

e) \(x^2+4\)

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

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

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

f) \(5x^2-7x-6\)

\(=\left(5x^2-10x\right)+\left(3x-6\right)\)

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

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

17 tháng 7 2019

a) =2x^3-10x^2-2x+3x^2-x

=2x^3-7x^2-3x

17 tháng 7 2019

b) -10x^4y^2z^2+35x^3y^2z^2+4x^4y^2z^2+4x^3y^2z^2

=-6x^4y^2z^2+39x^3y^2z^2

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}

19 tháng 10 2015

1) x4y2 + x2y4 + x4y+ x2y = (x4y2 + x2y4) + (x4y+ x2y5) = x2y2.(x+ y2) + x2y3.(x+ y2) = x2y2.(x2+ y2) (1 + y) = [xy.(x2 + y2)].[xy(1+y)]

=> x4y2 + x2y4 + x4y+ x2y5 chia cho xy.(x2 + y2)  bằng xy.(1+ y)

2) A = (n2 - 8)+ 36 = n4 - 16n2 + 100  = (n+ 20n2 + 100) - 36n= (n+ 10)- (6n)= (n2 - 6n+ 10).(n+ 6n+ 10)

Vậy để A là số nguyên tố thì n- 6n + 10 = 1 hoặc n+ 6n + 10 = 1

Mà n là số tự nhiên nên n2+ 6n + 10 > 1 

=>  n- 6n + 10 = 1  => n- 6n + 9 = 0 => (n -3)= 0 => n = 3 

Vậy....

3) a) = xy(x - y) - xz(x + z) + yz.[(x+ z) + (x - y)] = xy(x - y) - xz(x + z) + yz.(x + z) + yz(x - y)

= [xy(x - y) + yz.(x - y)] + [(yz.(x+ z) - xz(x+z)] = y(x - y)(x+ z) + z(x + z).(y - x) = (x+ z)(x- y).(y - z)

b) = (x+ x)- (2x)- 4(x+3) = (x+ x + 2x).(x+ x- 2x) - 4(x+3) = (x+ 3x).(x- x) - 4(x+3)

= (x+3).[x.(x2 - x) - 4] = (x+3).(x- x2 - 4) = (x+3).(x3 - 8 + 4 - x2) = (x+3).[(x - 2)(x2 + 2x + 4) - (x - 2).(x+2)]

= (x + 3).(x - 2).(x+ 2x + 4 - x- 2) = (x + 3).(x - 2).(x+ x + 2) 

4) a) n+ 1/4 = (n+ n+ 1/4) - n= (n+ 1/2)2 - n= (n2 - n + 1/2).(n+ n + 1/2) = [n(n - 1) + 1/2].[n.(n+1) + 1/2]

Áp dụng công thức ta có:

A = \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}=\frac{\frac{1}{2}.\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right)...\left(18.19+\frac{1}{2}\right).\left(19.20+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right).\left(4.5+\frac{1}{2}\right)...\left(19.20+\frac{1}{2}\right).\left(20.21+\frac{1}{2}\right)}\)

A = \(\frac{\frac{1}{2}}{20.21+\frac{1}{2}}=\frac{1}{841}\)