mn ơi  giúp em

Bài 3:

\(a,=3x\left(y-4x+6y^2\right)\\ b,=5xy\left(x^2-6x+9\right)=5xy\left(x-3\right)^2\\ d,=\left(x+y\right)\left(x-12\right)\\ f,=2x\left(x-y\right)\left(5x-4y\right)\\ g,=\left(x-2\right)\left(x-2+3x\right)=\left(x-2\right)\left(4x-2\right)=2\left(x-2\right)\left(2x-1\right)\\ h,=x^2\left(1-5x\right)+3xy\left(5x-1\right)=x\left(1-5x\right)\left(x-3y\right)\\ i,=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\\ j,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ k,=4x^2-12x+3x-9=\left(x-3\right)\left(4x+3\right)\\ l,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ m,=x^2-\left(2y-6\right)^2=\left(x-2y+6\right)\left(x+2y-6\right)\\ n,=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-25\\ =\left(x^2+5x\right)\left(x^2+5x+10\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

a) \(\dfrac{A}{x-3}=\dfrac{y-x}{3-x}\left(Đk:x\ne3\right)\)

\(A=\dfrac{\left(x-3\right)\left(y-x\right)}{3-x}=x-y\)

b) \(\dfrac{5x}{x+1}=\dfrac{Ax\left(x-1\right)}{\left(1-x\right)\left(x+1\right)}\left(Đk:x\ne\pm1\right)\)

\(A=\dfrac{5x\left(1-x\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-5\)

c) \(\dfrac{4x^2-5x+1}{A}=\dfrac{4x-1}{x+3}\left(Đk:x\ne-3;A\ne0\right)\)

\(A=\dfrac{\left(4x^2-5x+1\right)\left(x+3\right)}{4x-1}=\dfrac{\left(x-1\right)\left(4x-1\right)\left(x+3\right)}{4x-1}\)

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

tv cũ comback hả?

Chia câu ra được ko ạ

vấn đề chính giúp em còn nó không chia được

Câu 10:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{2;-1\right\}\\y\ne-5\end{matrix}\right.\)

\(A=\dfrac{y+5}{x^2-4x+4}\cdot\dfrac{x^2-4}{x+1}\cdot\dfrac{x-2}{y+5}\)

\(=\dfrac{y+5}{y+5}\cdot\dfrac{\left(x^2-4\right)}{x^2-4x+4}\cdot\dfrac{x-2}{x+1}\)

\(=\dfrac{\left(x^2-4\right)\cdot\left(x-2\right)}{\left(x+1\right)\left(x^2-4x+4\right)}\)

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

b: \(A=\dfrac{x+2}{x+1}\)

=>A không phụ thuộc vào biến y

Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+2\right):\left(\dfrac{1}{2}+1\right)=\dfrac{5}{2}:\dfrac{3}{2}=\dfrac{5}{2}\cdot\dfrac{2}{3}=\dfrac{5}{3}\)

Câu 12:

a: \(A=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\)

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

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

\(=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x-3}\)

b: Khi x=1 thì \(A=\dfrac{3}{1-3}=\dfrac{3}{-2}=-\dfrac{3}{2}\)

\(x+\dfrac{1}{3}=\dfrac{10}{3}\)

=>\(x=\dfrac{10}{3}-\dfrac{1}{3}\)

=>\(x=\dfrac{9}{3}=3\left(loại\right)\)

Vậy: Khi x=3 thì A không có giá trị

c: \(B=A\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x^2-4x+5}\)

\(x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1>=1\forall x\) thỏa mãn ĐKXĐ

=>\(B=\dfrac{3}{x^2-4x+5}< =\dfrac{3}{1}=3\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-2=0

=>x=2

Bài 1: 

a: =3x(x+2)

b: \(=x\left(x-1\right)^2\)

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

a) Ta có: AB//CD.

=>ABH=BDC (2 góc so le trong).

=> ∆AHB~∆BCD(g.g).

b) ∆ABD có :  DB²=AB²+AD²( Định lý Pitago)

=> DB= 15(cm).

Ta có ∆ABH~∆BCD(cmt).

=>AH/BC=AD/BD.

Hay AH=9.12/15=7,2(cm).

c)Ta có ∆AHB~∆BCD cmt.

=> HBA=CBD. (1)

Ta lại có : CBD= ADH (AB//CD).(2)

Từ 1 và 2 => HAB=ADH.

=>∆DHA~∆AHB(g.g).

S∆DHA/S∆AHB=(AD/AB)²=9/16

d) từ câu (a) và (b) => ∆BCD~∆DHA.

Cm ∆DHA~∆MDA(g.g)

Từ đó  suy ra ∆BDC~∆MDA.

Sau đó cm ∆BCD~∆ADC(g.g).

=> ∆MDA~∆ADC(g.g).

=>Ad/DC=DM/DC.

=>Đpcm.

Giúp e bài hình với ạ.

Bài 2: 

Xé ΔADH vuông tại H và ΔCBK vuông tại K có 

AD=BC

\(\widehat{ADH}=\widehat{CBK}\)

Do đó: ΔADH=ΔCBK

Suy ra: AH=CK

Xét tứ giác AHCK có 

AH//CK

AH=CK

Do đó: AHCK là hình bình hành

Bài 1:

1) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

2) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

3) \(\Rightarrow\left(4x-3\right)\left(7-12x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{7}{12}\end{matrix}\right.\)

4) \(\Rightarrow x^3+8-x^3+25x=-17\)

\(\Rightarrow25x=-25\Rightarrow x=-1\)

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

\(\Rightarrow\left(3x-2\right)\left(3x+2-6x+4\right)=0\)

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

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

Bài 3: 

c: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

d: \(x^3-7x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

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