1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

Bài 1 : 

\(\frac{4x-5}{x-1}=\frac{2+x}{x-1}\)ĐK : x \(\ne\)1

\(\Leftrightarrow\frac{4x-5}{x-1}-\frac{2-x}{x-1}=0\Leftrightarrow\frac{4x-5-2+x}{x-1}=0\)

\(\Rightarrow5x-7=0\Leftrightarrow x=\frac{7}{5}\)( tmđk )

Vậy tập nghiệm của phuwong trình là S= { 7/5 }

b, \(\frac{x-1}{x-2}-3+x=\frac{1}{x-2}\)ĐK : x \(\ne\)2

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

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

\(\Leftrightarrow\frac{x-1-3x+6+x^2-2x-1}{x-2}=0\)

\(\Rightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)( ktmđkxđ )

Vậy phương trình vô nghiệm 

c, \(1+\frac{1}{2+x}=\frac{12}{x^3+8}\)ĐK : x \(\ne\)-2 

\(\Leftrightarrow\frac{\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4-12}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Rightarrow x^3+8+x^2-2x+4-12=0\)

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

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

Vậy tập nghiệm của phương trình là S = { 0 ; 1 } 

d, đưa về dạng hđt 

Bài 2 : làm tương tự, chỉ khác ở chỗ mẫu số phức tạp hơn tí thôi 

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

x + 2/5 - x/3 = x - 12/15

=> 15x/15 + 6/15 - 5x/15 = 15x/15 - 12/15 (nhân cả hai vế với 15)

=> 15x - 5x - 18 = 15x - 12 (thực hiện các phép tính)

=> 10x = 6 (rút gọn)

=> x = 6/10 = 3/5

Vậy nghiệm của phương trình là x = 3/5.

\(x+\dfrac{2}{5}-\dfrac{x}{3}=x-\dfrac{12}{15}\)

\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{6}{15}-\dfrac{5x}{15}=\dfrac{15x}{15}-\dfrac{12}{15}\)

\(\Leftrightarrow15x+6-5x=15x-12\)

\(\Leftrightarrow15x-5x-15x=-6-12\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\left(x\ne1;x\ne3\right)\)

\(\Leftrightarrow\frac{x+5}{x-1}-\frac{x+1}{x-3}+\frac{8}{x^2-4x+3}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\frac{x^2-1}{\left(x-3\right)\left(x-1\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2+2x-15-x^2+1+8}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow2x-4=0\)

<=> 2x=4

<=> x=2 (tmđk)
Vậy x=2

b) \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\left(x\ne\pm2\right)\)

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

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-4}{x^2-4}=0\)

\(\Leftrightarrow\frac{x^2+3x+2-5x+10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-2x+2}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x+2=0

<=> -2x=-2

<=> x=1 (tmđk)
Vậy x=1

1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)

Suy ra: \(5x^2+3x-9=5x^2-5x\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(tm\right)\)

2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(6x=3x-15\)

\(\Leftrightarrow3x=-15\)

hay \(x=-5\left(loại\right)\)

2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)

Vậy pt vô nghiệm.

3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)

\(\Rightarrow -3(x+4)-(3-5x)=x-4\)

\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)

1/ x²-3x+2=0

⇒ (x²-2x)-(x-2)=0

⇒ x(x-2)-(x-2)=0

⇒ (x-1)(x-2)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2) x²-6x+5=0

⇒x²-6x+9-4=0

⇒(x²-6x+9)-2²=0

⇒(x-3)²-2²=0

⇒(x-3-2)(x-3+2)=0

⇒(x-5)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

3) 2x²+5x+3=0

⇒ (2x²+2x)+(3x+3)=0

⇒ 2x(x+1)+3(x+1)=0

⇒ (x+1)(2x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-1,5\end{matrix}\right.\)

4) x²-8x+15=0

⇒ (x²-8x+16)-1=0

⇒ (x-4)²-1²=0

⇒ (x-4-1)(x-4+1)=0

⇒ (x-5)(x-3)=0

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

5) x²-x-12=0

⇒ (x²-4x)+(3x-12)=0

⇒ x(x-4)+3(x-4)=0

⇒ (x-4)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

1: Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

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

2: Ta có: \(x^2-6x+5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

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

3: Ta có: \(2x^2+5x+3=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)

4: Ta có: \(x^2-8x+15=0\)

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

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

5: Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)