ĐKXĐ: $x\geq 5$đặt $b=\sqrt{x+1}=> x=b^2-1$ ; $a=\sqrt[4]{x-5}$ $(a,b\geq 0)$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
ĐKXĐ: $x\geq 5$đặt $b=\sqrt{x+1}=> x=b^2-1$ ; $a=\sqrt[4]{x-5}$ $(a,b\geq 0)$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: $x\geq 5$đặt $b=\sqrt{x+1}=> x=b^2-1$ ; $a=\sqrt[4]{x-5}$ $(a,b\geq 0)$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