Ta có ; x² - 11x + 24

= x² - 3x - 8x + 24

= x(x - 3) - (8x - 24)

= x(x - 3) - 8(x - 3)

= (x - 3)(x - 8)

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

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

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

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

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

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

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

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

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:

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

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

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

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

Cái này chưa học bt làm mấy câu

b. x^2 + 2x - 3

= x^2 + 3x - x - 3

= x ( x - 1 ) + 3 ( x - 1 )

= ( x + 3 ) ( x - 1 )

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

\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)

\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)

\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)

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

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

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

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

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)

đặt \(x^2+5x+5=t\)

\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)

            \(=t^2-1-24\)

            \(=t^2-25\)

            \(=\left(t-5\right)\left(t+5\right)\)

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

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

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

học tốt

d) 

Hướng dẫn :

(x+1)(x+2)(x+3)(x+4)-24

= x(x+5)(x^2+5x+10)

P/s: có gì vào trang web mathway.com viết đa thức vào rồi nhấn " factor " là ra nhân tử nhé

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

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

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

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

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

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

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

\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)

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

Tham khảo nhé~

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

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

lm tiếp câu c

c)  \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)

\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)

\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)

Đặt   \(x^2-9x+17=a\) ta có:

        \(C=\left(a-3\right)\left(a+3\right)-72\)

            \(=a^2-9-72\)

           \(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được:  \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)

tách nhỏ câu hỏi ra bạn

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

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

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)

=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)

=\(\left(3x-4\right).\left(x+14\right)\)