Giải phương trình vô tỉ Căn 3 (x-2)-Căn 3(2x-2)=-1 Căn 2 (x^2+x-1)-Căn 2(-x^2+x+1)=x^2-x+2 (Áp dụng côsi nhớ chứng minh) Căn 2 (x-4)+ Căn 2(6-x)=x^2

 1)$\sqrt[3]{x-2}$ +$\sqrt[3]{2x-2}$ =-1

($\sqrt[3]{x-2}$ +$\sqrt[3]{2x-2}$ )³=-1

x-2+2x-2+3$\sqrt[3]{(x-2)(2x-2)}$ ($\sqrt[3]{x-2}$ +$\sqrt[3]{2x-2}$)=-1

3x-4+1+3$\sqrt[3]{(x-2)(2x-2)}$ =0 // do $\sqrt[3]{x-2}$ +$\sqrt[3]{2x-2}$ =-1

x-1+$\sqrt[3]{(x-2)(2x-2)}$=0

x-1=-$\sqrt[3]{(x-2)(2x-2)}$

(x-1)³=(x-2)(2x-2)

x³-3.x²+3x-1=2.x²-4x-2x+4

x³-3x²+3x-1-2x²+4x+2x-4=0

x³-5x²+9x-5=0

(x-1)(x²-4x+5)=0

⇒x=1

2) đặt x²+x+1=a

⇔$\sqrt{a-2}$  $\sqrt{a}$  =a+1

⇔($\sqrt{a-2}$  $\sqrt{a}$)²=(a+1)²

⇔a-2+a+2$\sqrt{(a-2).a}$= a²+2a+1

⇔2$\sqrt{(a-2).a}$=a²+3

⇔(2$\sqrt{(a-2).a}$)²=(a²+3)²

4.(a-2)a=$a^{4}$ +6.a²+9

4.a²-8a=$a^{4}$ +6.a²+9

$a^{4}$ +6.a²+9-4a²+8a=0

$a^{4}$-2a²+8a+9=0

(a+1)(a³-a²-a+9)=0

tìm a thì xong