<=> \(\left(x-1\right)=\left(-2022\right)^2\)

<=> \(x-1=4088484\)

=> x = 4088484 + 1 = 4088485

Làm từ muối biển.

Anh chị tham khảo thử ạ:

Ta có: \(f\left(x\right)=\dfrac{2\sqrt{x}+m}{\sqrt{x+1}}\left(x\ge0\right)\)

\(\Rightarrow\)\(f'\left(x\right)=\dfrac{2.\dfrac{\sqrt{x+1}}{2\sqrt{x}}-\left(\sqrt{x}+m\right).\dfrac{1}{2\sqrt{x+1}}}{x+1}\)

\(\Rightarrow f'\left(x\right)=\dfrac{\dfrac{\sqrt{x+1}}{\sqrt{x}}-\dfrac{\sqrt{x}+m}{2\sqrt{x+1}}}{x+1}\)

1B

`B` nhé

tham khảo :

ĐKXĐ : \(x^2\ge y^2\)

P/t (2) <=> \(\dfrac{2y^2}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}=y\)  \(\Leftrightarrow\left[{}\begin{matrix}y=0\\\dfrac{2y}{\sqrt{x^2+y^2}+\sqrt{x^2-y^2}}=1\end{matrix}\right.\)

Với y = 0 thay vào p/t (1) : \(x^4=144\Leftrightarrow x=\sqrt[4]{144}\)

Với \(2y=\sqrt{x^2+y^2}+\sqrt{x^2-y^2}\)  .  Suy ra : \(\left\{{}\begin{matrix}\sqrt{x^2+y^2}=\dfrac{3y}{2}\\\sqrt{x^2-y^2}=\dfrac{y}{2}\end{matrix}\right.\)

Xét : \(\sqrt{x^2+y^2}=\dfrac{3y}{2}\Leftrightarrow x^2+y^2=\dfrac{9y^2}{4};y\ge0\)  \(\Leftrightarrow x^2=\dfrac{5y^2}{4}\) ; \(y\ge0\)

Thay vào p/t (1) : \(\dfrac{9y^2}{4}.\dfrac{1}{4}y^2=144\Leftrightarrow y^4=256\Leftrightarrow y=\pm4\) ; y \(\ge0\Rightarrow y=4\)

\(\Rightarrow x=...\)

\(x=234-12=222\)

\(x=123-12=111\)

12 + x = 234

x = 234 - 12

x = 222

123 - x = 12

x = 123 - 12 

x = 111.