Toán Lớp 9: a) (2x-1)(x+7)=x^2-49 b)2x+1/x-3 – x/x+3 =1 c)x+2/3-3x-1/5<-2

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1. 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
2. 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