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2 tháng 6 2018

\(a^5+b^5-\left(a+b\right)^5=a^5+b^5-\left(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\right)\)( tam giác Pascal )

\(=5a^4b+10a^3b^2+10a^2b^3+5ab^4=5ab\left(a^3+2a^2b+2ab^2+b^3\right)\)

\(=5ab\left(\left(a^3+b^3\right)+\left(2a^2b+2ab^2\right)\right)=5ab\left(\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right)\)

\(=5ab\left(a+b\right)\left(a^2+ab+b^2\right)\)

2 tháng 6 2018

Bạn làm đúng nhưng quên đấu âm

1 tháng 6 2018

a^3(c−b^2)+b^3(a−c^2)+c^3(b−a^2)+abc(abc−1)

=a^3c−a^3b^2+b^3(a−c^2)+bc^3−a^2c^3+a^2b^2c^2−abc

=(a^3c−a^2c^3)+b^3(a−c^2)−(a^3b^2−a^2b^2c^2)+(bc^3−abc)

=a^2c(a−c^2)+b^3(a−c^2)−a^2b^2(a−c^2)−bc(a−c^2)

=(a^2c+b^3−a^2b^2−bc)(a−c2)

=[c(a^2−b)−b^2(a^2−b)](a−c^2)=(a^2-b)(c-b^2)(a-c^2)

1 tháng 6 2018

Thanks

17 tháng 8 2018

\(\left(c^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left(a^2+b^2-2ab-9\right)\left(a^2+b^2+2ab-1\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

7 tháng 5 2022

`a)5(x-y)-y(x-y)`

`=(x-y)(5-y)`

`b)x^2-6x-y^2+9`

`=(x^2-6x+9)-y^2`

`=(x-3)^2-y^2`

`=(x-3-y)(x-3+y)`

(x+2).(x+3).(x+4).(x+5)−24

=(x2+7x+10).(x2+7x+12)−24

=(x2+7x+10).(x2+7x+10+2)−24

Đặt x2+7x+10=t, ta có

t.(t+2)−24

=t2+2t−24

=t2+2t+1−25

=(t−1)2−25

=(t−1−5)(t−1+5)

=(t−6)(t+4)

=(x2+7x+10−6)(x2+7x+10+4)

(x2+7x+4)(x2+7x+14)

P/s tham khảo nha

\(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)

\(\Leftrightarrow\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

\(\Leftrightarrow\left(x^2+7x+10\right).\left(x^2+7x+10+2\right)-24\)

Đặt \(x^2+7x+10=t\), ta có

\(t.\left(t+2\right)-24\)

\(\Leftrightarrow t^2+2t-24\)

\(\Leftrightarrow t^2+2t+1-25\)

\(\Leftrightarrow\left(t-1\right)^2-25\)

\(\Leftrightarrow\left(t-1-5\right)\left(t-1+5\right)\)

\(\Leftrightarrow\left(t-6\right)\left(t+4\right)\)

\(\Rightarrow\left(x^2+7x+10-6\right)\left(x^2+7x+10+4\right)\)

\(\Leftrightarrow\left(x^2+7x+4\right)\left(x^2+7x+14\right)\)

P/s tham khảo nha

1 tháng 11 2018

\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)

\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)

\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)

\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)

\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)

\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)

28 tháng 9 2017

a)hiệu hai lập phương

b) tổng hai lập phương

hằng đẳng thức có sẵn

28 tháng 9 2017

bạn giải ra hộ mk đi

28 tháng 7 2018

1.

\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

2.

a)  \(27x^4-8x=x\left(27x^3-8\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b)  \(16x^2y-4xy^2-4x^3+x^2y\)

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

\(=x\left(4x-y\right)\left(4y-x\right)\)

c) \(x^2-2x-5+2\sqrt{5}\)

\(=\left(x-1\right)^2-6+2\sqrt{5}\)

\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)

28 tháng 7 2018

Bài 1:

 \(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)

\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)

\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)

\(=5x^2-3xy^2+4y\)

Bài 2: 

a) \(27x^4-8x\)

\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)

\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)

b) \(16x^2y-4xy^2-4x^3+x^2y\)

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

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

31 tháng 5 2018

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a\left(a ^3+3.a^22b+3.a2b^2+2b^3\right)-b\left(2a^3+3.2a^2.b+3.2a.b^2+b^3\right)\)

\(=a\left(a^3+6a^2b+6ab^2+8b^3\right)-b\left(8a^3+6a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+6a^2b^2+8ab^3-8a^3b-6a^2b^2-6ab^3-b^4\)

\(=a^4+6a^3b+8ab^3-8a^3b-6ab^3-b^4\)

31 tháng 5 2018

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b\right)\left(a-b\right)^3\)