Bài 3: Phương trình đưa được về dạng ax + b = 0

Sau 30p xe 1 đi được 0,5*30=15km

=>Còn lại là 120-15=105km

Hai xe gặp nhau sau:

105/(30+45)=105/75=7/5(giờ)

\(\dfrac{x+3}{x-3}+\dfrac{x-3}{x+3}=\dfrac{x^2+2x}{x^2-9}+1\left(ĐLXĐ:x\ne3;x\ne-3\right)\)

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

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

\(\Leftrightarrow x^2+3x+3x+9+x^2-3x-3x+9=x^2+2x\)

\(\Leftrightarrow x^2+x^2-x^2+3x+3x-3x-3x-2x+9+9=0\)

\(\Leftrightarrow-2x-18=0\)

\(\Leftrightarrow-2x=18\)

\(\Leftrightarrow x=-9\left(nhận\right)\)

Vậy \(S=\left\{-9\right\}\)

ĐKXĐ:\(x\ne-2\)

\(\dfrac{1}{x+2}-1=\dfrac{5x+7}{x+2}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{5x+7}{x+2}=1\\ \Leftrightarrow\dfrac{1-5x-7}{x+2}=1\\ \Leftrightarrow-5x-6=x+2\\ \Leftrightarrow x+2+5x+6=0\\ \Leftrightarrow6x+8=0\\ \Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

:v

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

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

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

\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Tách nhỏ câu hỏi ra bạn

Bài 1 : 

a, đk x khác -1 ; 0 

\(x+1-2x=3\Leftrightarrow x=-2\)(tm) 

b, đk : x khác 0 ; 3/2 

\(x-3=10x-15\Leftrightarrow9x=12\Leftrightarrow x=\dfrac{4}{3}\)(tm) 

c, đk x khác 0 ; 2 

\(x^2+2x-x+2=2\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0\left(ktm\right);x=-1\)

d, đk : x khác -2 ; 2 

\(x^2-4x+4-3x-6=2x-22\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow x=4;x=5\)

Bài 1:

a) \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\) (1)

ĐKXĐ: \(x\ne0;x\ne-1\)

\(\left(1\right)\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{2x}{x\left(x+1\right)}=\dfrac{3}{x\left(x+1\right)}\)

\(\Leftrightarrow x+1-2x=3\)

\(\Leftrightarrow-x=2\)

\(\Leftrightarrow x=-2\) (nhận)

Vậy S = {\(-2\)}

45 phút = \(\dfrac{3}{4}h.\)

Gọi quãng đường AB là x (km) \(\left(x>0\right).\)

Thời gian đi từ A đến B là \(\dfrac{x}{48}\left(h\right).\)

Thời gian đi về là \(\dfrac{x}{48+12}=\dfrac{x}{60}\left(h\right).\)

Vì thời gian về ít hơn thời gian đi là 45 phút, nên ta có PT:

\(\dfrac{x}{60}+\dfrac{3}{4}=\dfrac{x}{48}.\\ \Rightarrow x=180\left(TM\right).\)

\(\Leftrightarrow x\cdot\dfrac{7}{8}-5x+45=\dfrac{10}{3}x+\dfrac{1}{4}\)

=>-179/24x=-179/4

hay x=6

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)