a, gọi MN cắt BC tại O 

xét tam giácBDC có : M là trung điểm của BD (gt)

MO // DC (Gt)

=> O là trung điểm của BC  (đl)

xét tam giác ABC có : NO // AB

=> N là trung điểm của AB (đl)

Hình :

\(\frac{x^2+3x}{x-1}.Q=\frac{x^2-9}{x^2-x}\)

\(\Leftrightarrow Q=\frac{x^2-9}{x\left(x-1\right)}:\frac{x\left(x+3\right)}{x-1}\)

\(\Leftrightarrow Q=\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-1\right)}.\frac{x-1}{x\left(x+3\right)}\)

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

Ta có : \(\frac{x^2+3x}{x-1}.Q=\frac{x^2-9}{x^2-x}\)

\(\Leftrightarrow\frac{x\left(x+3\right)}{x-1}.Q=\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-1\right)}\)

\(\Leftrightarrow Q=\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-1\right)}:\frac{x\left(x+3\right)}{x-1}\)

\(\Leftrightarrow Q=\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-1\right)}\cdot\frac{x-1}{x\left(x+3\right)}\)

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

Vậy : \(Q=\frac{x-3}{x^2}\)

a)\(\frac{2x-1}{x-1}-\frac{1}{x-1}=-1\)1

\(2=-1\)(vô lý)

Vậy phương trình này vô nghiệm

Các câu này phải có điều kiện xác định ( ĐKXĐ ) Bạn tự tìm nhé :

a) pt \(\Leftrightarrow\frac{2x-1-1+x-1}{x-1}=0\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\) ( loại do không thoa mãn ĐKXĐ )

b) pt \(\Leftrightarrow\frac{5x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)+12\left(x+1\right)}{2\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow5x^2-5x+2x^2-2+12x+12=0\)

\(\Leftrightarrow7x^2+7x-10=0\)

\(\Leftrightarrow x=\frac{-7\pm\sqrt{329}}{14}\) ( thoả mãn )

c) d) Tương tự

a) ĐKXĐ : \(x\ne-5\)

Pt \(\Leftrightarrow2x-5=3\left(x+5\right)\)

\(\Leftrightarrow x=-20\) ( thỏa mãn ĐKXĐ )

b) ĐKXĐ : \(x\ne0\)

Pt  \(\Leftrightarrow x^2-6=x\left(x+\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{3}{2}x=-6\Leftrightarrow x=-4\)

...

\(c,\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

\(\Leftrightarrow x^2+2x-3x+6=0\)

\(\Leftrightarrow x^2-x+6=0\)

\(\Leftrightarrow Vô-nghiệm\left(\Delta1-6=-5< 0\right)\)

Vậy pt vô nghiệm

\(d,\frac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(2x-1\right)\left(3x+2\right)\)

\(\Leftrightarrow6x^2+4x-3x-2=5\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+\frac{7}{6}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-7}{6}\end{cases}}\)

Vậy ............

Thêm đk nha bạn

\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)

Vậy .........

\(b,\left(x^2-4\right)+\left(x-2\right)\left(3-2x=0\right)\)

\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)

\(\Leftrightarrow-x^2+7x-10=0\)

\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)

Vậy ..................

\(c,x^3-3x^2+3x-1=0\)

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

\(\Leftrightarrow x=1\)

\(d,x\left(2x-7\right)-4x+14=0\)

\(\Leftrightarrow2x^2-7x-4x+14=0\)

\(\Leftrightarrow2x^2-11x+14=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)

Vậy ............

\(e,\left(2x-5\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)

\(\Leftrightarrow3x^2-24x+21=0\)

\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy .....................

\(f,x^2-x-\left(3x-3\right)=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy ..............

a, 8/x-8 + 11/x-11 = 9/x-9  + 10/ x-10

b, x/x-3 - x/x-5 = x/x-4 - x/x-6

c, 4/x^2-3x+2  - 3/2x^2-6x+1   +1 = 0

d, 1/x-1 + 2/ x-2  + 3/x-3  = 6/x-6

e, 2/2x+1 - 3/2x-1 = 4/4x^2-1

f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3

g, 1/x-1 + 2x^2 -5/x^3 -1  = 4/ x^2 +x+1

\(a,\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

Vậy ............

\(b,\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2,3x=6,9\\0,1x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

Vậy ...........

\(c,\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-0,5\\x\in\varnothing\end{cases}}\)

Vậy .........................

\(d,\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+7=0\\x-5=0\\5x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-7\\x=5\\5x=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{cases}}\)

Vậy ...............

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

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\)

\(\Leftrightarrow x=-\frac{1}{2}\) ( do \(x^2+1\ge1>0\forall x\) )

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-5=0\end{cases}hoặc5x+1=0}\)

\(\Leftrightarrow x\in\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}\)

a) \(7+2x=22-3x\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)

b) \(8x-3=5x+12\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

c) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow3x=36\Leftrightarrow x=12\)

d) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow3x=24\Leftrightarrow x=8\)

e) \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow x=7\)

f) \(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow0x=9\) ( vô lí )