Bài 1
`1) 5x(x+3)=5x²-30`
`⇔5x²+15x=5x²-30`
`⇔5x²+15x-5x²=-30`
`⇔15x=-30`
`⇔x=-30:15=-2`
Vậy `S={-2}`
$2)\dfrac{5x-2}{6}+\dfrac{3-4x}{2}=2-\dfrac{x+7}{3}$
$⇔\dfrac{5x-2}{6}+\dfrac{3(3-4x)}{6}=2-\dfrac{2(x+7)}{6}$
`⇔5x-2+3(3-4x)=2-2(x+7)`
`⇔5x-2+9-12x=2-2x-14`
`⇔5x-12x+2x=2-14+2-9`
`⇔-5x=-19`
`⇔x=19/5`
Vậy `S={19/5}`
`3)| 2x-3| =5`
⇒$\left \{ {{2x-3=5 khi x\geq0} \atop {-2x+3=5khi x<0}} \right.$
⇔$\left \{ {{2x=8 } \atop {-2x=2}} \right.$
⇔$\left \{ {{x=4(nhận) } \atop {x=-1(nhận)}} \right.$
Vậy `S={4;-1}`
$4)\dfrac{x-5}{x-1}+\dfrac{2}{x-3}=1$
ĐKXĐ: `x ne 1 và x ne 3`
$⇔\dfrac{(x-5)(x-3)}{(x-1)(x-3)}+\dfrac{2(x-1)}{(x-3)(x-1)}=\dfrac{(x-3)(x-1)}{(x-3)(x-1)}$
`⇒(x-5)(x-3)+2(x-1)=(x-3)(x-1)`
`⇔x^2-3x-5x+15+2x-2=x^2-x-3x+3`
`⇔x^2-3x-5x+2x+3x-x^2+x=3-15+2`
`⇔-2x=-10`
`⇔x=5(nhận)`
Vậy `S={5}`
Bài 2
`1)(2x-1)^2+7>x(4x+3)+1`
`⇔4x^2-4x+1+7>4x^2+3x+1`
`⇔4x^2-4x-4x^2-3x>1-1-7`
`⇔-7x> -7`
`⇔x<1`
Vậy bất phương trình có tập nghiệm {$x/x<1$}
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0 1
$2)\dfrac{12x+1}{12} ≥ \dfrac{9x+3}{3}-\dfrac{8x+1}{4}$
$⇔\dfrac{12x+1}{12} ≥ \dfrac{4(9x+3)}{12}-\dfrac{3(8x+1)}{12}$
`⇔12x+1≥36x+12-(24x+1)`
`⇔12x+1≥36x+12-24x-1`
`⇔12x-36x+24x≥12-1-1`
`text{⇔0x≥10(vô nghiệm)}`
Vậy bất phương trình có tập nghiệm $∅$
