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}$$

@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