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25 tháng 11 2021

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

\(=2x^3+5x^2+6x-4x^2-10x-12\)

\(=\left(2x^3+5x^2+6x\right)-\left(4x^2+10x+12\right)\)

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

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

 

25 tháng 11 2021

\(a,2x^3+x^2-4x-12=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)

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

\(c,\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]-9=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)

đặt \(x^2+8x+11=y\)

\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9=\left(y-4\right)\left(y+4\right)-9=y^2-16-9=y^2-25=\left(y-5\right)\left(y+5\right)=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)=\left(x^2+8x+6\right)\left(x^2+8x+16\right)=\left(x^2+8x+6\right)\left(x+4\right)^2\)

Ói , hoa mắt chóng mặt nhức đầu ,

9 tháng 8 2017

sao giống có chữa quá z

a: \(15xy^2z^3:3xyz^2=5yz\)

b: \(12x^4y^4:\left(-4x^4y^2\right)=-3y^2\)

c: \(\dfrac{-15x^2y^3z^2}{-6xz^2}=\dfrac{5}{2}xy^3\)

d: \(\dfrac{\left(x-y\right)^5}{\left(y-x\right)^3}=-\left(x-y\right)^2\)

21 tháng 9 2021

mình có 6ý mà

12 tháng 10 2021

a) \(x^2+4x+4-y^2\)

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

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

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

12 tháng 10 2021

\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)

24 tháng 10 2017

b; 2x3+x2-4x-12

= 2x3-4x2+5x2-10x+6x-12

=(2x3-4x2)+(5x2-10x)+(6x-12)

=2x2(x-2)+ 5x(x-2)+6(x-2)

= (x-2)(2x2+5x

24 tháng 10 2017

con a) mik cứ thấy sai sai nên bn xem lại nhé:

b) 2x3 + x2 - 4x - 12

= 2x3 - 4x2 + 5x2 - 10x + 6x - 12

= (2x3 - 4x2) + (5x2 - 10x) + (6x - 12)

= 2x2(x - 2) + 5x(x - 2) + 6(x - 2)

= (2x2 + 5x + 6)(x - 2)

= (2x2 + 2x + 3x + 6)(x - 2)

= [(2x2 + 2x) + (3x + 6)](x - 2)

= [2x(x + 2) + 3(x + 2)](x - 2)

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

= (2x + 3)(x2 - 4)

c) x3 - 9x2 + 14x

= x3 - 7x2 - 2x2 + 14x

= (x3 - 7x2) - (2x2 - 14x)

= x2(x - 7) - 2x(x - 7)

= (x2 - 2x)(x - 7)

= x(x - 2)(x - 7)

d) x5 - xy4 + x4y - y5

= (x5 + x4y) - (xy4 + y5)

= x4(x + y) - y4(x + y)

= (x4 - y4)(x + y)

= (x2 + y2)(x2 - y2)(x + y)

= (x2 + y2)(x - y)(x + y)(x + y)

= (x2 + y2)(x - y)(x + y)2

e) (x + 1)(x + 3)(x + 5)(x + 7) - 9

= [(x + 1)(x + 7)][(x + 3)(x + 5)] - 9

= (x2 + 7x + x + 7)(x2 + 5x + 3x + 15) - 9

= (x2 + 8x + 7)(x2 + 8x + 15) - 9

Thay x2 + 8x + 7 = y, ta có:

y(y + 8) - 9

= y2 + 8y - 9

= y2 - y + 9y - 9

= (y2 - y) + (9y - 9)

= y(y - 1) + 9(y - 1)

= (y + 9)(y - 1)

= (x2 + 8x + 7 + 9)(x2 + 8x + 7 - 1)

= (x2 + 8x + 16)(x2 + 8x + 6)

= (x + 4)2.(x2 + 8x + 6)

NHỚ TIK MK NHÉ

okokok

Bài 4 :

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

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

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

d)

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

e) Trùng câu d

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

Bài 5:

a) \(x^3-x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\

\(\Leftrightarrow2x-3=6\)

\(\Leftrightarrow x=\frac{9}{2}\)

vậy........

c) \(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)

Vậy

d) \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy ........

Bài 1 : 

a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)

TH1 : \(x^2-2x+3=0\)

\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm 

TH2 : \(x-4=0\Leftrightarrow x=4\)

b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)

TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)

\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)

TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)

c, đưa về hệ đc ko ? 

d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)

TH1 : \(x=0,74...\) ( bấm máy cx ra )

TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm 

KL : vô nghiệm 

Bài 2 : 

a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)

\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)

Vậy biểu thức ko phụ thuộc vào biến 

b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)

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

\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến