Câu 1:
`a)5(x+2)=3x-10`
`⇔5x+10=3x-10`
`⇔5x-3x=-10-10`
`⇔2x=-20`
`⇔x=(-20):2`
`⇔x=-10`
Vậy `S={-10}`
`b)(2x+5)(3x-1)=(2x+5)(x-3)`
`⇔(2x+5)(3x-1)-(2x+5)(x-3)=0`
`⇔(2x+5)[(3x-1)-(x-3)]=0`
`⇔(2x+5)(3x-1-x+3)=0`
`⇔(2x+5)(2x+2)=0`
`⇔`$\left[\begin{matrix} 2x+5=0\\ 2x+2=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 2x=-5\\ 2x=-2\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-\dfrac{5}{2}\\ x=-1\end{matrix}\right.$
Vậy `S={-5/2;-1}`
`c)2/(x²+2x)-1/x=(x-2)/(x+2)(ĐKXĐ:x`$\neq$ `0,x`$\neq$ `-2)`
`⇔2/[x(x+2)]-1/x=(x-2)/(x+2)`
`⇔2/[x(x+2)]-(x+2)/[x(x+2)]=[x(x-2)]/[x(x+2)]`
`⇒2-(x+2)=x(x-2)`
`⇔2-x-2=x²-2x`
`⇔2-x-2-x²+2x=0`
`⇔-x²+(-x+2x)+(2-2)=0`
`⇔-x²+x=0`
`⇔-x(x-1)=0`
`⇔`$\left[\begin{matrix} -x=0\\ x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0(Ko TM ĐKXĐ)\\ x=1(TM ĐKXĐ)\end{matrix}\right.$
Vậy `S={1}`
`d)|2x-1|+5=6`
`⇔|2x-1|=6-5`
`⇔|2x-1|=1`
`⇔`$\left[\begin{matrix} 2x-1=1\\ 2x-1=-1\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 2x=2\\ 2x=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=1\\ x=0\end{matrix}\right.$
Vậy `S={0;1}`
`e)5(x-3)=3x+10`
`⇔5x-15=3x+10`
`⇔5x-3x=10+15`
`⇔2x=25`
`⇔x=25/2`
Vậy `S={25/2}`
`f)(2x-5)(3x+1)=(2x-5)(x+3)`
`⇔(2x-5)(3x+1)-(2x-5)(x+3)=0`
`⇔(2x-5)[(3x+1)-(x+3)]=0`
`⇔(2x-5)(3x+1-x-3)=0`
`⇔(2x-5)(2x-2)=0`
`⇔`$\left[\begin{matrix} 2x-5=0\\ 2x-2=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 2x=5\\ 2x=2\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\dfrac{5}{2}\\ x=1\end{matrix}\right.$
Vậy `S={5/2;1}`
`g)2/(x²-2x)+1/x=(x+2)/(x-2)(ĐKXĐ:x`$\neq$ `0,x`$\neq$ `2)`
`⇔2/[x(x-2)]+1/x=(x+2)/(x-2)`
`⇔2/[x(x-2)]+(x-2)/[x(x-2)]=[x(x+2)]/[x(x-2)]`
`⇒2+x-2=x(x+2)`
`⇔2+x-2=x²+2x`
`⇔2+x-2-x²-2x=0`
`⇔-x²+(x-2x)+(-2+2)=0`
`⇔-x²-x=0`
`⇔-x(x+1)=0`
`⇔`$\left[\begin{matrix} -x=0\\ x+1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0(Ko TM ĐKXĐ)\\ x=-1(TM ĐKXĐ)\end{matrix}\right.$
Vậy `S={-1}`
`h)|2x-1|-5=4`
`⇔|2x-1|=4+5`
`⇔|2x-1|=9`
`⇔`$\left[\begin{matrix} 2x-1=9\\ 2x-1=-9\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 2x=10\\ 2x=-8\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=5\\ x=-4\end{matrix}\right.$
Vậy `S={5;-4}`
Câu 2:
`a)3-2x≥4`
`⇔-2x≥4-3`
`⇔-2x≥1`
`⇔x≤-1/2`
Vậy `S={x|x≤-1/2}`
Biểu diễn trên trục số:`(ảnh 1)`
`b)(4x-5)/3>(7-x)/5`
`⇔[5(4x-5)]/15>[3(7-x)]/15`
`⇔5(4x-5)>3(7-x)`
`⇔20x-25>21-3x`
`⇔20x+3x>21+25`
`⇔23x>46`
`⇔x>46:23`
`⇔x>2`
Vậy `S={x|x>2}`
Biểu diễn trên trục số:`(ảnh 1)`
`c)4x-3≥2x+4`
`⇔4x-2x≥4+3`
`⇔2x≥7`
`⇔x≥7/2`
Vậy `S={x|x≥7/2}`
Biểu diễn trên trục số:`(ảnh 1)`
`d)(-2x-3)/4>(x-4)/3`
`⇔[3(-2x-3)]/12>[4(x-4)]/12`
`⇔3(-2x-3)>4(x-4)`
`⇔-6x-9>4x-16`
`⇔-6x-4x>``-16+9`
`⇔-10x>``-7`
`⇔x<7/10`
Vậy `S={x|x<7/10}`
Biểu diễn trên trục số:`(ảnh 1)`
