Kết quả là 25

P không có max bạn nhé. Tìm được min thôi.

Lời giải:

Có: \(P=\frac{\sqrt{x}+1}{2-\sqrt{x}}=\frac{\sqrt{x}+1}{2-\sqrt{x}}+1-1=\frac{3}{2-\sqrt{x}}-1\)

Do $\sqrt{x}\geq 0$ với mọi $x$

$\Rightarrow 2-\sqrt{x}\leq 2$

$\Rightarrow P\geq \frac{3}{2}-1=\frac{1}{2}$

Vậy $P_{\min}=\frac{1}{2}$. Giá trị này đạt tại $x=0$

\(D=\dfrac{2\left(\sqrt{x}-1\right)+9}{\sqrt{x}-1}=2+\dfrac{9}{\sqrt{x}-1}\)

Vì \(\dfrac{9}{\sqrt{x}-1}\le\dfrac{9}{0-1}=-9\Leftrightarrow D\le2-9=-7\)

Vậy \(D_{max}=-7\Leftrightarrow x=0\)

1/ \(y'=\dfrac{\left(\sqrt{x+1}\right)'x-x'\sqrt{x+1}}{x^2}=\dfrac{\dfrac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}=\dfrac{-x-2}{2x^2\sqrt{x+1}}\)

2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)

3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)

4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)

5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học

ok cảm ơn bạn nhìu

a: ĐKXĐ: \(\begin{cases}x\ge0\\ x<>1\end{cases}\)

b: \(x=4-2\sqrt3\)

=>\(x=\left(\sqrt3-1\right)^2\)

=>\(\sqrt{x}=\left|\sqrt3-1\right|=\sqrt3-1\)

\(f\left(x\right)=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt3-1+1}{\sqrt3-1-1}=\frac{\sqrt3}{\sqrt3-2}=\frac{\sqrt3\left(\sqrt3+2\right)}{-1}=-\sqrt3\left(2+\sqrt3\right)=-2\sqrt3-3\)

\(f\left(a^2\right)=\frac{\sqrt{a^2}+1}{\sqrt{a^2}-1}=\frac{\left|a\right|+1}{\left|a\right|-1}=\frac{-a+1}{-a-1}=\frac{a-1}{a+1}\)

c: f(x)=f(x^2)

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

=>\(x+1=\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)=x+2\sqrt{x}+1\)

=>\(2\sqrt{x}=0\)

=>x=0(nhận)