Question

a)\(\left(x^2+\frac{4}{x^2}\right)-4\left(x-\frac{2}{x}\right)-9=0\)

b)\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)

Answer 1

a) x≠0

x^2+4/x^2=(x-2/x)^2+4

Pt<=>(x-2/x)^2+4-4(x-2/x)-9=0

<=>(x-2/x)^2-4(x-2/x)-5=0

Đặt t=x-2/x

Pt<=> t^2-4t-5=0

Đến đây tìm t rồi quy đồng lên tìm ra x nhé!

Answer 2

b)x>=-2

(√(x+5)-√(x+2))(1+√(x^2+7x+10))=3

<=> (√(x+5)-√(x+2))(1+√(x+5)(x+2))=3

Đặt √(x+5)=a;√(x+2)=b (a>b>=0)

=> a^2-b^2=3

Pt<=>(a-b)(1+ab)=a^2-b^2

<=>(a-b)(1+ab)=(a-b)(a+b)

Mà a>b=>a-b>0

=>ab+1=a+b

<=>(a-1)(b-1)=0

a=1=>x+5=1<=>x=-4(loại)

b=1=>x+2=1<=>x=-1 (thoả mãn)

Vậy x=-1