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
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
By Quỳnh Ngân
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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>43/4Vậy S={x|x>43/4}
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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)=1ĐKXĐ : 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>43/4Vậy S={x|x>43/4}