Bài 1: Mở đầu về phương trình

4) \(\Rightarrow\left(3x+7\right)\left(x+5\right)-\left(x+5\right)\left(3-2x\right)=0\)

\(\Rightarrow\left(x+5\right)\left(5x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{4}{5}\end{matrix}\right.\)

5) \(\Rightarrow2x\left(x-3\right)\left(x+3\right)-5\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+3\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: =>(x+2)(x-1)=0

=>x=-2 hoặc x=1

3: =>(x-3)(x+3-2x-1)=0

=>(x-3)(2-x)=0

=>x=3 hoặc x=2

\(2.x^2+x-2=0\\ \Rightarrow x^2+2x-x-2=0\\ \Rightarrow x\left(x+2\right)-\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\\ 3.x^2-9-\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left[\left(x+3\right)-\left(2x+1\right)\right]=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-1\right)=0\\ \Rightarrow\left(x-3\right)\left(-x+2\right)=0\\ \Rightarrow\left(x-3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

  2) x²+x-2=0

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

3) x²-9-(2x+1)(x-3)=0

    x²-9-2x²+6x+x-3=0

    (x²-2x²)+6x+x-3-9=0

    -x²+7x-12=0

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

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

(x2+5x)2−2(x2+5x)−24=[(x2+5x)2−6(x2+5x)]+[4(x2+5x)−24]=(x2+5x)(x2+5x−6)+4(x2+5x−6)=(x2+5x−6)(x2+5x+4)=(x2−x+6x−6)(x2+4x+x+4)=[x(x−1)+6(x−1)]+[x(x+4)+(x+4)]=(x−1)(x+1)(x+4)(x+6)

a Xét tứ giác ADME có 

\(\widehat{ADM}=\widehat{AEM}=\widehat{DAE}=90^0\)

Do đó: ADME là hình chữ nhật

a: Xét tứ giác BMDN có 

DM//BN

DM=BN

Do đó: BMDN là hình bình hành

\(C=8+6+10=24\left(cm\right)\)

b: Xét tứ giác ABCM có

F là trung điểm của AC
F là trung điểm của BM

Do đó: ABCM là hình bình hành

Suy ra: AM//BC

Bài 3:

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