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a) Để hàm xác định thì \(\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b) Ta có: \(f\left(x\right)=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(\Rightarrow f\left(4-2\sqrt{3}\right)=\frac{\sqrt{4-2\sqrt{3}}+1}{\sqrt{4-2\sqrt{3}}-1}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{\sqrt{\left(\sqrt{3}-1\right)^2}-1}=\frac{\sqrt{3}}{\sqrt{3}-2}\)
và \(f\left(a^2\right)=\frac{\sqrt{a^2}+1}{\sqrt{a^2}-1}=\frac{\left|a\right|+1}{\left|a\right|-1}\)(với \(a\ne\pm1\))
* Nếu \(a\ge0;a\ne1\)thì \(f\left(a^2\right)=\frac{a+1}{a-1}\)
* Nếu \(a< 0;a\ne-1\)thì \(f\left(a^2\right)=\frac{a-1}{a+1}\)
c) \(f\left(x\right)=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để f(x) nguyên thì \(\frac{2}{\sqrt{x}-1}\)nguyên hay \(2⋮\sqrt{x}-1\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Mà \(\sqrt{x}-1\ge-1\)nên ta xét ba trường hợp:
+) \(\sqrt{x}-1=-1\Rightarrow x=0\left(tmđk\right)\)
+) \(\sqrt{x}-1=1\Rightarrow x=4\left(tmđk\right)\)
+) \(\sqrt{x}-1=2\Rightarrow x=9\left(tmđk\right)\)
Vậy \(x\in\left\{0;4;9\right\}\)thì f(x) có giá trị nguyên
d) \(f\left(x\right)=\frac{\sqrt{x}+1}{\sqrt{x}-1}\); \(f\left(2x\right)=\frac{\sqrt{2x}+1}{\sqrt{2x}-1}\)
f(x) = f(2x) khi \(\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{2x}+1}{\sqrt{2x}-1}\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{2x}+1\right)\)\(\Leftrightarrow\sqrt{2}x+\sqrt{2x}-\sqrt{x}-1=\sqrt{2}x-\sqrt{2x}+\sqrt{x}-1\)\(\Leftrightarrow\sqrt{2x}-\sqrt{x}=-\sqrt{2x}+\sqrt{x}\Leftrightarrow2\sqrt{2x}=2\sqrt{x}\Leftrightarrow\sqrt{2x}=\sqrt{x}\Leftrightarrow x=0\)(tmđk)
Vậy x = 0 thì f(x) = f(2x)
a, ĐKXĐ : x > 0 và x khác 9
F = x-\(3\sqrt{x}\)+x+\(3\sqrt{x}\)/x-9 . x-9/\(\sqrt{x}\)
= 2x/x-9 . x-9/\(\sqrt{x}\) = 2x\(\sqrt{x}\)
b, F = 1/2 <=> 2x\(\sqrt{x}\)=1/2
<=>x\(\sqrt{x}\) = 1/4 hay \(\sqrt{x}^3\) = 1/4
<=> \(\sqrt{x}=\sqrt[3]{\frac{1}{4}}\)
<=> x=\(\sqrt[3]{\frac{1}{4}}^2\)
Nếu đúng thì k mk nha
a) TXĐ:\(x\ge0\)
b)\(f\left(4-2\sqrt{3}\right)=\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}\)\(=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}}=\frac{3-2\sqrt{3}}{3}\)
\(f\left(a^2\right)=\frac{\left(-a\right)-1}{\left(-a\right)+1}=\frac{-1-a}{1-a}\)
c)\(f\left(x\right)\in Z\Rightarrow1-\frac{2}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x\in\left\{0;1\right\}TM\)
d)\(f\left(x\right)=f\left(x^2\right)\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\left|x\right|-1}{\left|x\right|+1}=\frac{x-1}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(\sqrt{x}-1\right)=\left(x-1\right)\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow-x+\sqrt{x}=x-\sqrt{x}\)
\(\Rightarrow x=0;1\)(TM)
+KL...
#Walker
a: \(f\left(x\right)=\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)
\(f\left(-1\right)=\left|-1-3\right|=4\)
\(f\left(5\right)=\left|5-3\right|=\left|2\right|=2\)
b: f(x)=10
=>\(\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)
c: \(A=\dfrac{f\left(x\right)}{x^2-9}=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\)
TH1: x<3 và x<>-3
=>\(A=\dfrac{-\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\)
TH2: x>3
\(A=\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)