View attachment 154099
Mình xin cám ơn trước ạ
[tex]x+1=\sqrt{2(x+1)+2\sqrt{2(x+1)+4\sqrt{x+1}}}\\\\ <=> x^2-1= 2\sqrt{2(x+1)+4\sqrt{x+1}}\\\\ +, x \neq -1; 1 \\\\ x+1=\sqrt{2(x+1)+2\sqrt{2(x+1)+4\sqrt{x+1}}}=\frac{2(x+1)+2\sqrt{2(x+1)+4\sqrt{x+1}}}{x+1}\\\\ => x-1= \frac{4.\sqrt{2(x+1)+4\sqrt{x+1}}}{x+1}=\frac{4.[2(x+1)+4\sqrt{x+1}]}{(x^2-1).(x+1)}\\\\ => (x-1)^2.(x+1)^2=4(x+1)+8\sqrt{x+1}\\\\ => (a^2-2)^2.a^4=8a^2+16a\\\\ <=> a. [a^3(a^4-4a^2+4)-8a-16]=0\\\\ <=> a^7-4a^5+4a^3-8a-16=0\\\\ <=> a^7-2a^6+2a^6-4a^5+4a^3-8a^2+8a^2-16a+8a-16=0\\\\
<=> (a-2).(a^6+2a^5+4a^2+8a+8)=0\\\\
<=> a-2=0 (a>0) <=> a=2\\\\
<=>a^2=4 => x+1=4 <=> x=3 [/tex]
với x=-1 => thỏa mãn
với x=1 => ko thỏa mãn
vậy ....
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