Bài 1 tìm x: a, $\sqrt[]{-x^2+x+4 }$ =x-3 b, $\sqrt[]{x-3}$ -2$\sqrt[]{x^2-9}$ =0 Bài 2: tìm y biết $\sqrt[]{4y-20}$ +$\sqrt[]{y-5}$ -$\frac{1}{3}$$

0 bình luận về “Bài 1 tìm x: a, $\sqrt[]{-x^2+x+4 }$ =x-3 b, $\sqrt[]{x-3}$ -2$\sqrt[]{x^2-9}$ =0 Bài 2: tìm y biết $\sqrt[]{4y-20}$ +$\sqrt[]{y-5}$ -$\frac{1}{3}$$”

1. @nt0950925

   Bài 1:

   a/ -x²+x+4=x²-6x+9

   <=> 2x²-7x+5=0

   <=> (x-1)(2x-5)=0

   <=>\(\left[ \begin{array}{l}x1=1\\x2=5/2\end{array} \right.\) 

   b/ $\sqrt[]{x-3}$ =2$\sqrt[]{x^{2}-9}$

   <=>$\sqrt[]{x-3}$ =2$\sqrt[]{x-3}$$\sqrt[]{x+3}$

   TH1: $\sqrt[]{x-3}$=0 <=> x=3

   TH2: 2.$\sqrt[]{x+3}$ =1 <=>$\sqrt[]{x+3}$ =1/2 <=>x=-11/4

   Bài 2:

   $\sqrt[]{(y-5).4}$+$\sqrt[]{y-5}$-1/3.$\sqrt[]{(y-5).9}$=4

   <=>2.$\sqrt[]{(y-5)}$+$\sqrt[]{y-5}$ -$\sqrt[]{(y-5)}$=4

   <=>2.$\sqrt[]{(y-5)}$=4

   <=>$\sqrt[]{y-5}$=2

   <=>y-5=4

   <=>y=9

   Bình luận