a) x^2-4x=0 b)(x+4)(7x-3)-x^2+16=0 Giups em với ak

2 bình luận về “a) x^2-4x=0 b)(x+4)(7x-3)-x^2+16=0 Giups em với ak”

1. a) x^2-4=0

   <=> x(x-4)=0

   <=>\(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\) 

   Vậy x=0 hoặc x=4

   ———-

   b) (x+4)(7x-3)-x^2 +16=0

   <=> (x+4)(7x-3)-(x^2 -16)=0

   <=> (x+4)(7x-3)-(x+4)(x-4)=0

   <=> (x+4)(7x-3-x+4)=0

   <=> (x+4)(6x+1)=0

   <=> \(\left[ \begin{array}{l}x+4=0\\6x+1=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=-4\\x=-\dfrac{1}{6}\end{array} \right.\) 

   Vậy x=-4 hoặc x=-1/6

   ~~Hok tốt~~

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2. @Rwyz

   a) x^2-4x=0

   <=> x(x-4)=0

   <=> $\left[\begin{matrix} x=0\\ x-4=0\end{matrix}\right.$

   <=> $\left[\begin{matrix} x=0\\ x=4\end{matrix}\right.$

   Vậy x=0 hoặc x=4

   b) (x+4)(7x-3)-x^2+16=0

   <=> (x+4)(7x-3)-x^2-4^2=0

   <=> (x+4)(7x-3)-(x+4)(x-4)=0

   <=> (x+4)[(7x-3)-(x-4)] =0

   <=> (x+4)[7x-3-x+4]=0

   <=> (x+4)[6x+1]=0

   <=> $\left[\begin{matrix} x+4=0\\ 6x+1=0\end{matrix}\right.$

   <=> $\left[\begin{matrix} x=-4\\ x=-1/6\end{matrix}\right.$

   Vậy x=-4 hoặc x=-1/6

    

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