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18 tháng 2 2021

 a) 3x2 – 7x + 2

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

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

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

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

 b) a(x2 + 1) – x(a2 + 1)

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

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

.......?

 

 

 

 

a) Ta có: \(3x^2-7x+2\)

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

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

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

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

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

1 tháng 11 2021

\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)

1 tháng 11 2021

bạn giải lại giúp mình bài 2 được ko ạ

 

17 tháng 7 2021

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

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

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

 

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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

22 tháng 11 2020

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

17 tháng 11 2019

a) (x - 1)(x + l)(x - 2)(x - 4).      b) (x - 2)( x 2  + 4).

c) 2y(3 x 2   +   y 2 ).                          d) 2(x + y + z) ( a   -   b ) 2 .

24 tháng 8 2021

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

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

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

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

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

b. \(x^3-2x^2+4x-8\)

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

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

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

c. \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

\(=6x^2y+2y^3\)

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

d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)

\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)

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

20 tháng 7 2021

a) x2-4y2-x++2y

= x2-(2y)2-x+2y

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

=(x-2y)(x+2y-1)

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)

\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)

hay P=x+3

15 tháng 9 2021

1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\)

2) \(9-6\sqrt{a}+a=\left(\sqrt{a}-3\right)^2\)

3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)

4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)

5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)

15 tháng 9 2021

1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)

2) \(9-6\sqrt{a}+a=\left(3-\sqrt{a}\right)^2\)

3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)

4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)

5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1^2=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)

23 tháng 8 2021

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

23 tháng 8 2021

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)