ĐKXĐ: x≥5đặt b=√x+1=>x=b2−1 ; a=4√x−5 (a,b≥0)pt<=>$a^4+\frac{2(b^2-1)a}{b+1}=b^2$$a^4(b+1)+2a(b+1)(b-1)-b^2(b+1)=0$$(b+1)(a^4+2a(b-1)-b^2)=0=>b=-1(KTĐK)$=> pt đã cho vô nghiệm
ĐKXĐ:
x≥5đặt
b=√x+1=>x=b2−1 ;
a=4√x−5 (a,b≥0)pt<=>$a^
2+\frac{2(b^2-1)a}{b+1}=b^2$$a^
2(b+1)+2a(b+1)(b-1)-b^2(b+1)=0$$(b+1)(a^
2+2a(b-1)-b^2)=0
=>b=-1(KTĐK)$=> pt đã cho vô nghiệm