\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)

\(\Leftrightarrow\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)

\(\Rightarrow\left(x-2\right)^2=x^2-4\)

\(\Leftrightarrow x^2-4x+4-x^2+4=0\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow x=2\)

Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{x-2}=b;\) ta có:

\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)

\(\Leftrightarrow2a^2+ab-2ab-b^2=0\)

⇔ \(\left(2a+b\right)\left(a-b\right)=0\)

⇔ \(\left[{}\begin{matrix}2\sqrt[3]{x+2}=-\sqrt[3]{x-2}\\\sqrt[3]{x-2}=\sqrt[3]{x+2}\end{matrix}\right.\)⇔ \(x=-\frac{14}{9}\)

\(\Leftrightarrow\dfrac{x+3+x-1+2\sqrt{\left(x+3\right)\left(x-1\right)}}{x+3-x+1}=\dfrac{13-x^2}{4}\)

\(\Leftrightarrow2x+2+2\sqrt{\left(x+3\right)\left(x-1\right)}=13-x^2\)

\(\Leftrightarrow\sqrt{4\left(x+3\right)\left(x-1\right)}=13-x^2-2x-2=-x^2-2x+11\)

=>\(x\simeq1,37\)

\(\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3\)

\(\Leftrightarrow2x+2\sqrt{\left(x-\sqrt{2-x}\right)\left(x+\sqrt{x-2}\right)}=9\)

\(\Leftrightarrow2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x+2}\right)}=9-2x\)

\(\Leftrightarrow4\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)=\left(9-2x\right)^2\)

\(\Leftrightarrow4x^2-4x+8=81-36x+4x^2\)

\(\Leftrightarrow-4x+8=81-36x\)

\(\Leftrightarrow-4x=81-36x-8\)

\(\Leftrightarrow-4x=-36x+73\)

\(\Leftrightarrow-4x+36x=73\)

\(\Leftrightarrow32x=73\)

\(\Leftrightarrow x=\frac{73}{32}\)

Vậy: nghiệm phương trình là: \(\left\{\frac{73}{32}\right\}\)

Lỗi sai ngu người nhất của Chihiro.Quên viết ĐKXĐ ak em

\(\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3\)

\(ĐKXĐ:x\ge2\)

Bình phương 2 vế của pt ta được

\(2x+2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)}=9\)

\(\Leftrightarrow2\sqrt{x^2-x+2}=9-2x\)

\(\Leftrightarrow\hept{\begin{cases}9-2x\ge0\Leftrightarrow\frac{9}{2}\ge x\\4\left(x^2-x+2\right)=81-36x+4x^2\left(2\right)\end{cases}}\)

\(\left(2\right)\Leftrightarrow32x-73=0\Leftrightarrow x=\frac{73}{32}\left(tmDK\right)\)

Vậy \(S=\left\{\frac{73}{32}\right\}\)

p/s:học hỏi đi con.

Không thích thì không ghi được không ạ? :))

sáng sớm lang thang lật lại mấy trang gặp bài này, xin trình bày vài cách:

Đk:\(x\ge2\) \(\left(DK\forall PP\right)\)

C1 \(pt\Leftrightarrow x^3-3x\left(x+2\right)-2\sqrt{\left(x+2\right)^3}=0\)

Đặt \(t=\sqrt{x+2}\) ra pt đăng cấp bậc 3...

c2:\(pt\Leftrightarrow\left(\sqrt{\left(x+2\right)^3}+1\right)^2=\left(3\left(x+1\right)\right)^2\)

c3:\(pt\Leftrightarrow\left(\sqrt{\left(x+2\right)^3}-3x-2\right)\left(3x+\sqrt{\left(x+2\right)^3+4}\right)=0\)

C4:Chia 2 vế x³ dc:

\(1-\frac{3}{x}\pm2\sqrt{\left(\frac{1}{x}+\frac{2}{x^2}\right)}-\frac{6}{x^2}=0\)

đặt \(\sqrt{\left(\frac{1}{x}+\frac{2}{x^2}\right)}=t\) dc \(1\pm3t^2+2t^3=0\)

Ngoài ra còn có thể liên hợp ,.....

a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)

b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)

c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)

Đặt \(\sqrt{x^2+3}=a\ge\sqrt{3}\) (1)

pt \(\Leftrightarrow\left(a^2-3\right)^2+a-3=0\)

\(\Leftrightarrow a^4+9-6a^2+a-3=0\)

\(\Leftrightarrow a^4-4a^2-2a^2+4a-3a+6=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2-2a-3=0\right)\)

\(\Leftrightarrow\left(a-2\right)\left(a^3+a^2+a^2+a-3a-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left(a^2+a-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left[\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}a-2=0\\a+1=0\\\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(c\right)\\a=-1\left(l\right)\\a=\dfrac{-1+\sqrt{13}}{2}\left(l\right)\\a=\dfrac{-1-\sqrt{13}}{2}\left(l\right)\end{matrix}\right.\)

Thay a = 2 vào (1) ta được: \(\sqrt{x^2+3}=2\Rightarrow x^2+3=4\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy ...

Vây phương trình có nghiêm là x=1 hay x=-1