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31 tháng 10 2020

a) Đặt: x = a- b; y = b - c ; z = c- a 

Ta có: x + y + z = 0 

=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)

=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

b) Đặt: \(a=x^2-2x\) 

Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)

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

d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)

Đặt: \(x^2-8=t\)

Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)

\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)

\(=\left(2x^2+9x-16\right)^2\)

4 tháng 7 2016

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

\(x^2+5x+4=x^2+4x+x+4=x\left(x+4\right)+\left(x+4\right)=\left(x+1\right)\left(x+4\right)\)

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

\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2\right)^2-\left(2x^2\right)=\left(x^2+2+2x\right)\left(x^2-2-2x\right)\)

4 tháng 7 2016

Nhần mình ấn lộn bài sory m.n

24 tháng 9 2021

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

8 tháng 2 2019

Bài 1 :

\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)

\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)

\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)

Đặt \(a=x^2+6x-7\)

\(A=a\left(a-9\right)+8\)

\(A=a^2-9a+8\)

\(A=a^2-8a-a+8\)

\(A=a\left(a-8\right)-\left(a-8\right)\)

\(A=\left(a-8\right)\left(a-1\right)\)

Thay a vào là xong bạn :)

cảm ớn phương nhiều

3 tháng 9 2021

\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2\left(x^2-2x+1\right)-5x+5-1\right)=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)

3 tháng 9 2021

\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)

\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left[2\left(x^2-2x+1\right)-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)

\(=\left(x-1\right)\left(2x^2-9x+6\right)\)

28 tháng 8 2023

1) \(3x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(3x+5\right)\)

2) \(4x(x-2y)-8y(2y-x)\)

\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)

\(=\left(4x+8y\right)\left(x-2y\right)\)

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

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

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

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

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

4) \(3x\left(x-a\right)+4a\left(a-x\right)\)

\(=3x\left(x-a\right)-4a\left(x-a\right)\)

\(=\left(x-a\right)\left(3x-4a\right)\)

5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)

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

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

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

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

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

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

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

7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)

\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)

\(=\left(x-y\right)\left(a-m\right)^2\)

8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)

\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)

\(=\left(6y^2+9x\right)\left(x-1\right)^2\)

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

#Ayumu

29 tháng 9 2016
X^4+x^2+1 X^4+2x^2+1_x^2 =(x^2+1)^2-x^2 Hang dang thuc 3 =
29 tháng 9 2016

1

(x2-8)2+36

=x4-16x2+64+36

=x4+20x2+100-36x2

=(x2+10)2-(6x)2

HĐT số 3

20 tháng 7 2018

a) 1/2(x3+8)=1/2(x+2)(x2-2x+4)

b) x4(x-y)+2x3(x-y)=x3(x+2)(x-y)

c) x2-(y2-6y+9)=x2-(y-3)2=(x-y+3)(x+y-3)

d) xy(x3+y3)=xy(x+y)(x2-xy+y2)

e)3x2(x2-25y2)=3x2(x-5y)(x+5y)

f) 4x4+4x2y2+y4-4x2y2= (2x2+y2)2-(2xy)2=(2x2-2xy+y2)(2x2+2xy+y2)

20 tháng 7 2018

a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)

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

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

e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).

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

28 tháng 10 2016

Đăng từng câu thôi như hế này thì chắc .....

28 tháng 10 2016

mình sẽ đăng từ từ thôi nên mong bạn giải giúp mình với.mình cảm ơn