a)(2x-1)(x+7)=x²-49 ⇔(2x-1)(x+7)=(x+7)(x-7) ⇔(2x-1)(x+7)-(x+7)(x-7)=0 ⇔(x+7)[(2x-1)-(x-7)]=0 ⇔(x+7)(2x-1-x+7)=0 ⇔(x+7)(x+6)=0 ⇔\(\left[ \begin{array}{l}x+7=0\\x+6=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-7\\x=-6\end{array} \right.\) Vậy S={-7;-6} b)(2x+1)/(x-3)-x/(x+3)=1(ĐKXĐ:x$\neq$ ±3) ⇔[(2x+1)(x+3)]/[(x+3)(x-3)]-[x(x-3)]/[(x+3)(x-3)]=[(x+3)(x-3)]/[(x+3)(x-3)] ⇒(2x+1)(x+3)-x(x-3)=(x+3)(x-3) ⇔2x²+6x+x+3-x²+3x=x²-9 ⇔(2x²-x²)+(6x+x+3x)+3=x²-9 ⇔x²+10x+3=x²-9 ⇔x²+10x-x²=-9-3 ⇔10x=-12 ⇔x=-12/10 ⇔x=-6/5(TM ĐKXĐ) Vậy S={-6/5} c)(x+2)/3-(3x-1)/5<-2 ⇔[5(x+2)]/15-[3(3x-1)]/15<-30/15 ⇒5(x+2)-3(3x-1)<-30 ⇔5x+10-9x+3<-30 ⇔5x-9x<-30-10-3 ⇔-4x<-43 ⇔x>43/4 Vậy S={x|x>43/4} Trả lời
Giải đáp + Lời giải và giải thích chi tiết: a) (2x-1)(x+7)=x^2-49 <=> (2x-1)(x+7)-(x^2-49)=0 <=> (2x-1)(x+7)-(x-7)(x+7)=0 <=> (x+7)(2x-1-x+7)=0 <=> (x+7)(x+6)=0 <=> \(\left[ \begin{array}{l}x+7=0\\x+6=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=-7\\x=-6\end{array} \right.\) Vậy S={-7;-6} b) (2x+1)/(x-3)-x/(x+3)=1 ĐKXĐ : x ne +-3 <=> ((2x+1)(x+3)-x(x-3))/(x^2-9)=(x^2-9)/(x^2-9) => (2x+1)(x+3)-x(x-3)=x^2-9 <=> 2x^2+6x+x+3-x^2+3x-x^2+9=0 <=> 10x+12=0 <=> 10x=-12 <=> x=-6/5 \ \ (tmđk) Vậy S={-6/5} c) (x+2)/3-(3x-1)/5<-2 <=> (5.(x+2)-3.(3x-1))/15< -30/15 => 5.(x+2)-3.(3x-1)=-30 <=> 5x+10-9x+3<-30 <=> -4x<-30-3-10 <=> -4x<-43 <=> x>43/4 Vậy S={x|x>43/4} Trả lời
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