B1:

\(a,A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

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

\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)

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

\(=\frac{3x+9-x^2-3x+x^2-3x}{x^2-9}.\frac{x+3}{3x^2}\)

\(=\frac{9-3x}{x^2-9}.\frac{x+3}{3x^2}\)

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

\(=\frac{3-x}{x^3-3x^2}\)

B2: 

\(a,B=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)

\(=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x+2}{x^2-4}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\left(\frac{x-2x-4+x-2}{x^2-4}\right):\frac{6}{x+2}\)

\(=-\frac{6}{x^2-4}.\frac{x+2}{6}\)

\(=\frac{-6\left(x+2\right)}{\left(x+2\right)\left(x-2\right)6}=-\frac{1}{x-2}\)

\(2|3-2x|-3x=13\)

* Nếu \(3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow x\le\frac{3}{2}\)

\(\Leftrightarrow|3-2x|=3-2x\)

\(2\left(3-2x\right)-3x=13\)

\(\Leftrightarrow6-4x-3x=13\)

\(\Leftrightarrow-7x-7=0\)

\(\Leftrightarrow x=-1\)( thỏa mãn )

* Nếu \(3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\frac{3}{2}\)

\(\Leftrightarrow|3-2x|=-3+2x\)

\(2\left(-3+2x\right)-3x=13\)

\(\Leftrightarrow-6+4x-3x=13\)

\(\Leftrightarrow x=19\) ( thỏa mãn )

Vậy phương trình có tập nghiệm \(S=\left\{-1;19\right\}\)

a) Xét  \(\Delta ABC\)có  \(AE=EB\)

                                  \(AD=DC\)

\(\Rightarrow\)ED là đường trung bình  \(\Delta ABC\)

\(\Rightarrow\hept{\begin{cases}ED=\frac{1}{2}BC\Leftrightarrow ED=\frac{1}{2}\times8=4\left(cm\right)\\ED//BC\end{cases}}\)

\(\Rightarrow\)EDCB là hình thang

Lại có :  \(EM=MB\)

             \(DN=NC\)

\(\Rightarrow\)MN là đường trung bình của hình thang EDCB

\(\Rightarrow MN=\frac{ED+BC}{2}=\frac{4+8}{2}=\frac{12}{2}=6\left(cm\right)\)

Vậy  \(MN=6cm\)

b) Xét  \(\Delta BED\)có M là trung điểm BE ; MI // ED

\(\Rightarrow\)MI là dường trung bình  \(\Delta BED\)

\(\Rightarrow MI=\frac{1}{2}ED=\frac{1}{2}\times4=2\left(cm\right)\)

Xét  \(\Delta CED\)có N là trung điểm CD ; NK // ED

\(\Rightarrow\)NK là đường trung bình  \(\Delta CED\)

\(\Rightarrow NK=\frac{1}{2}ED=\frac{1}{2}\times4=2\left(cm\right)\)

Lại có :  \(MI+IK+KN=MN\)

\(\Leftrightarrow2+IK+2=6\)

\(\Leftrightarrow IK=2\left(cm\right)\)

Vậy  \(MI=IK=KN\left(=2cm\right)\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x+8=15\)

\(\Leftrightarrow2x=15-8=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x.\left(x^2-2\right)=15\)

mị thấy cũng đơn giản mà sao lại hỏi

Mị tự học ở nhà, mà bài này lớp 8 (chưa học đến) hơi không hiểu nên mới hỏi :3

\(x^4-2x^2y^2+y^4-1=0\Leftrightarrow\left(x^2-y^2\right)^2-1=0\Leftrightarrow\left(x^2-y^2-1\right).\left(x^2-y^2+1\right)=0\\ \)

\(x^2+2xy+2x+2y+y^2+1=0\Leftrightarrow\left(x+y+1\right)^2=0\)

\(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\Leftrightarrow x^2+xy-2xy-2y^2=0\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Mà \(x+y\ne0\Rightarrow x-2y=0\Rightarrow x=2y\)

\(\Rightarrow A=\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)

x²  - 2y² = xy <=> x² - xy - 2y² = 0 <=> x² + xy - 2xy - 2y² = 0 <=> x (  x + y ) - 2y 

( x + y ) = 0 <=> ( x - 2y ) ( x + y ) = 0

mà x + y \(\ne\) 0 => x - 2y = 0 => x = 2y

=> A = \(\frac{2y-y}{2y+y}\) = \(\frac{y}{3y}\) = \(\frac{1}{3}\)

\(4x^2-x-\frac{3}{16}\)

\(=\left(2x\right)^2-x+\frac{1}{4}-\frac{7}{16}\)

\(=\left(2x-\frac{1}{2}\right)^2-\frac{7}{16}\)

Mà  \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{1}{2}\right)^2-\frac{7}{16}\ge-\frac{7}{16}\)

Dấu " = " xảy ra \(\Leftrightarrow\left(2x-\frac{1}{2}\right)^2=0\)

\(x=\frac{1}{4}\)

Vậy GTNN của biểu thức bằng \(-\frac{7}{16}\) tại \(x=\frac{1}{4}\)

Gọi biểu thức trên là A. Ta có:

\(A=4x^2-x-\frac{3}{16}\)

\(A=4x^2-2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-\frac{3}{16}\)

\(A=\left(2x-\frac{1}{2}\right)^2-\frac{1}{4}-\frac{3}{16}\)

\(A=\left(2x-\frac{1}{2}\right)^2-\frac{7}{16}\)

Nhận xét: \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{1}{2}\right)^2-\frac{7}{16}\ge\frac{-7}{16}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{4}\)

Vậy \(minA=\frac{-7}{16}\Leftrightarrow x=\frac{1}{4}\)