\(DK:x\le\sqrt[8]{17}\)

\(\Leftrightarrow\left(\sqrt[4]{17-x^8}-2\right)+\left(1-\sqrt[3]{2x^8-1}\right)=0\)

\(\Leftrightarrow\frac{\sqrt{17-x^8}-4}{\sqrt[4]{17-x^8}+2}+\frac{2\left(1-x^8\right)}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^2}=0\)

\(\Leftrightarrow\frac{1-x^8}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}+4\right)}+\frac{2\left(1-x^8\right)}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^2}=0\)

\(\Leftrightarrow\left(1-x^8\right)\left[\frac{1}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}\right)}+\frac{1}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)}\right]=0\)

Vi \(\frac{1}{\left(\sqrt[4]{17-x^8}+2\right)\left(\sqrt{17-x^8}\right)}+\frac{2}{1+\sqrt[3]{2x^8-1}+\left(\sqrt[3]{2x^8-1}\right)^8}>0\left(\forall x\le\sqrt[8]{17}\right)\)

\(\Rightarrow x^8=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=-1\left(n\right)\end{cases}}\)

Vay nghiem cua PT la \(x=-1\)

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

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

\(\Leftrightarrow\left(2+m\right)\left(2x-mx+2\right)=0\)

Xet voi \(m=-2\)thi PT tro thanh

\(4x+4=4x+4\)

\(\Leftrightarrow0x=0\)

Vay PT co vo so nghiem

Xet \(m\ne-2\)thi PT co nghiem la 

\(2x-mx+2=0\)

\(\Leftrightarrow\left(2-m\right)x=-2\)

\(\Leftrightarrow x=\frac{2}{m-2}\)

Suy ra: PT co nghiem khi \(m\ne2\)

Vay PT co nghiem tong quat la \(x=\frac{2}{m-2}\left(m\ne2\right)\)

là sao

Môn tin học lớp 10 ý bạn..

Bạn lớp mấy?.😪😪

Em kiểm tra lại đề bài

\(A=\left(-\infty;m+1\right)\)hay là \(A=\)( \(-\infty;m+1\)]