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

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)

b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)

\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)

B2:

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

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

Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv

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

b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)

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

\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)

\(=3b\left(2a-b\right)\)

`a, 4x^2-1 = (2x+1)(2x-1)`

`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`

`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`

Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

Đặt \(x^2+7x+10=y\)

\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)

\(A=y^2+2y+1-25\)

\(A=\left(y+1\right)^2-5^2\)

\(A=\left(y+1-5\right)\left(y+1+5\right)\)

\(A=\left(y-4\right)\left(y+6\right)\)

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

\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

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

Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=a\)

\(\Rightarrow B=a.\left(a+3\right)-4\)

\(B=a^2+3a-4\)

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

\(B=a.\left(a-1\right)+4.\left(a-1\right)\)

\(B=\left(a-1\right)\left(a+4\right)\)

\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

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

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

\(\Leftrightarrow2.2x=4x\)

p/s tham khảo nha

\(a^2-b^2-a+b\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)\)

p/s tham khảo

\(x^4+x^3+x+1\)

\(\Leftrightarrow x\left(x^3+1\right)+x^3+1\)

\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)\)

P/s tham khảo nha

1.

\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)

\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)

Hay đa thức trên có thể phân tích thành:

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

Dựa vào đó em tự tách cho phù hợp

2.

\(VT=a\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+b\left(\dfrac{1}{a^2}+\dfrac{1}{c^2}\right)+c\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)

\(VT\ge\dfrac{2a}{bc}+\dfrac{2b}{ac}+\dfrac{2c}{ab}=2\dfrac{a^2+b^2+c^2}{abc}\)

\(VP=\dfrac{2\left(ab+bc+ca\right)}{abc}\)

\(\Rightarrow\dfrac{ab+bc+ca}{abc}\ge\dfrac{a^2+b^2+c^2}{abc}\)

\(\Rightarrow ab+bc+ca\ge a^2+b^2+c^2\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\le0\)

\(\Rightarrow a=b=c\)

3.

\(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\)

\(\Rightarrow\left(\dfrac{x^2-yz}{a}\right)^2=\left(\dfrac{y^2-xz}{b}\right)\left(\dfrac{z^2-xy}{c}\right)=\dfrac{\left(x^2-yz\right)^2-\left(y^2-xz\right)\left(z^2-xy\right)}{a^2-bc}\)

\(=\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}\)

Tương tự:

\(\left(\dfrac{y^2-xz}{b}\right)^2=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}\)

\(\left(\dfrac{z^2-xy}{c}\right)^2=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x}{a^2-bc}=\dfrac{y}{b^2-ac}=\dfrac{z}{c^2-ab}\Rightarrowđpcm\)

a: =(x-3)(2x+5)

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

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...

a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)

\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)

\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)

\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)

b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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

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

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

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

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

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

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

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

d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)

\(=\left(a+1\right)\left(a^4+a^2+1\right)\)

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

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

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

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

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

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

f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)

\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)

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

g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)

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

a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

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

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

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

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

Giải giúp bạn 2 bài tiêu biểu thôi nha

không biết

a) x³ + (a+b+c)x²+ (ab+ac+bc)x +abc

= x³ +ax²+bx²+cx²+abx+acx+bcx+abc

=x³+cx²+abx+abc+ax²+acx+bx²+bcx

=x² (x+c) + ab (x+c) +ax (x+c) +bx (x+c)

= (x+c) (x²+ab+ax+bx)

= (x+c) { x(x+b)+a(x+b)}

=(x+c) (x+b) (x+a)