Không biết bạn có gõ đúng đề cả 2 câu không ? Câu 2 không có nghiệm nguyên dương nhé bạn. Bạn xem lại.

có đúng đề không bạn

Sửa lại đề \(\frac{x+1}{x-1}+\frac{x-2}{x+2}+\frac{x-3}{x+3}+\frac{x+4}{x-4}=-4\)

ĐK \(x\ne\left\{1;-2;-3;4\right\}\)

\(\Leftrightarrow\left(\frac{x+1}{x-1}+1\right)+\left(\frac{x-2}{x+2}+1\right)+\left(\frac{x-3}{x+3}+1\right)+\left(\frac{x+4}{x-4}+1\right)=0\)

\(\Leftrightarrow\frac{2x}{x-1}+\frac{2x}{x+2}+\frac{2x}{x+3}+\frac{2x}{x-4}=0\)

\(\Leftrightarrow2x\left(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\right)=0\Leftrightarrow x=0\)vì \(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\ne0\)

Vậy pt có nghiệm  \(x=0\)

cô giáo in đề cho mk là =4 mà

nếu k thì mk xong lâu r

\(\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}\)

\(\Leftrightarrow\sqrt{\frac{1-x}{x}}-1=\frac{2x+x^2}{1+x^2}-1\)

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

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

Dễ thấy: \(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}< 0\)

\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)

\(\frac{X+1}{99}+1+\frac{X+2}{98}+1+\frac{x+3}{97}+1+\frac{X+4}{96}+1=0\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{X+100}{98}+\frac{X+100}{97}+\frac{X+100}{96}=0\Leftrightarrow\left(X+100\right)\times\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0 \)\(\Leftrightarrow X+100=0\Leftrightarrow x=-100\)

ĐKXĐ: \(x>0\)

\(\Leftrightarrow x+\frac{1}{x}-4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+5=0\)

Đặt \(\sqrt{x}+\frac{1}{\sqrt{x}}=t>0\Rightarrow t^2=x+\frac{1}{x}+2\Rightarrow x+\frac{1}{x}=t^2-2\)

Pt trở thành:

\(t^2-2-4t+5=0\Leftrightarrow t^2-4t+3=0\) \(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{\sqrt{x}}=1\\\sqrt{x}+\frac{1}{\sqrt{x}}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}+1=0\left(vn\right)\\x-3\sqrt{x}+1=0\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)