Giải pt:

\(\left|x^3-x^2-x-2\right|=2x^3+2x^2+x-2\)

a)\(\dfrac{\sqrt{7}+5+\sqrt{7}-5}{\sqrt{7}-5}=\dfrac{2\sqrt{7}}{\sqrt{7}-5}=\dfrac{-7-5\sqrt{7}}{9}\approx-2,25\)

Tìm x hợ:

\(x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0\) hoặc \(x+1=0\)

TH1: x = 0

TH2: x = -1

Vậy x = 0, x= -1

a) \(4\sqrt{5}-\sqrt{50}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-\sqrt{50}\)

c) \(3\sqrt{3}+4\sqrt{2}-15\sqrt{3}=-12\sqrt{3}+4\sqrt{2}\)

Chậc, không biết đúng không nữa

e)\(\left(2+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right)\left(2-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\)

d) \(\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}+1-1}=\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}-1=0\)

Tính hả bạn?