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a: f(a)=g(a)
=>5a-3=-1/2a+1
=>5,5a=4
=>\(a=\dfrac{4}{5.5}=\dfrac{8}{11}\)
b: f(b-2)=g(2b+4)
=>\(5\left(b-2\right)-3=-\dfrac{1}{2}\left(2b+4\right)+1\)
=>\(5b-13=-b-2+1=-b-1\)
=>6b=12
=>b=2
f(a) = g(a)
⇔ 5a - 3 = -a/2 + 1
⇔ 5a + a/2 = 1 + 3
⇔ 11a/2 = 4
⇔ 11a = 8
⇔ a = 8/11
Vậy a = 8/11 thì f(a) = g(a)
b) f(b - 2) = g(2b + 4)
⇔ 5.(b - 2) - 3 = -(2b + 4)/2 + 1
⇔ 5b - 10 - 3 = -b - 2 + 1
⇔ 5b + b = 1 + 13
⇔ 6b = 14
⇔ b = 7/3
Vậy b = 7/3 thì f(b - 2) = g(2b + 4)
\(a,f\left(-3\right)=9;f\left(-\dfrac{1}{2}\right)=\dfrac{1}{4};f\left(0\right)=0\\ g\left(1\right)=2;g\left(2\right)=1;g\left(3\right)=0\\ b,2f\left(a\right)=g\left(a\right)\\ \Leftrightarrow2a^2=3-a\\ \Leftrightarrow2a^2+a-3=0\\ \Leftrightarrow2a^2-2a+3a-3=0\\ \Leftrightarrow2a\left(a-1\right)+3\left(a-1\right)=0\\ \Leftrightarrow\left(2a+3\right)\left(a-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{3}{2}\end{matrix}\right.\)
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)
b: Ta có: \(2\cdot f\left(a\right)=g\left(a\right)\)
\(\Leftrightarrow2a^2=3-a\)
\(\Leftrightarrow2a^2+a-3=0\)
\(\Leftrightarrow2a^2+3a-2a-3=0\)
\(\Leftrightarrow\left(2a+3\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{3}{2}\end{matrix}\right.\)
Câu 1: Sai vì nếu như f(x)=ax+b(b<>0) thì f(-a)=-ax+b<>ax+b
Câu 2:
a: f(x)=0
=>-2x+1/2=0
=>-2x=-1/2
hay x=1/4
b: g(x)=-7
=>3x-1/4=-7
=>3x=-27/4
hay x=-9/4