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Bài 1:
Ta có: xy ≤ (x + y)²/4 = 1/4, dấu = xảy ra khi x = y = 1/2
P = (x² + 1/y²)(y² + 1/x²) = (xy)² + 1 + 1 + 1/(xy)²
= (xy)² + 1/[256(xy)²] + 255/[256(xy)²] + 2
ta có:
(xy)² + 1/[256(xy)²] ≥ 2 √(1/256) = 1/8. dấu = xảy ra khi x = y = 1/2
255/[256(xy)²] + 2 ≥ 255/(256.1/16) + 2 = 287/16. dấu = xảy ra khi x = y = 1/2
cộng theo vế → P ≥ 1/8 + 287/16 = 289/16
vậy GTNN của P là 289/16, đạt được khi x = y = 1/2
a: ĐKXĐ: \(\left\{{}\begin{matrix}-2< =x< =2\\x< >0\end{matrix}\right.\)
c: \(f\left(-x\right)=\dfrac{\sqrt{2-\left(-x\right)}-\sqrt{2+\left(-x\right)}}{-x}=\dfrac{\sqrt{2+x}-\sqrt{2-x}}{-x}=\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}=f\left(x\right)\)
a: ĐKXĐ: (x+4)(x-1)<>0
hay \(x\notin\left\{-4;1\right\}\)
b: \(y-3=\dfrac{2x^2+6\sqrt{\left(x^2+1\right)\left(x-2\right)}+5-3x^2-9x+12}{x^2+3x-4}\)
\(=\dfrac{-x^2-9x+17+6\sqrt{\left(x^2+1\right)\left(x-2\right)}}{x^2+3x-4}< =0\)
=>y<=3
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
\(f\left(2k-1\right)=\left[\left(2k-1\right)^2+2k-1+1\right]^2+1\)
\(=\left(4k^2+1-2k\right)^2+1=\left(4k^2+1\right)^2-4k\left(4k^2+1\right)+4k^2+1\)
\(=\left(4k^2+1\right)\left(4k^2-4k+2\right)=\left(4k^2+1\right)\left[\left(2k-1\right)^2+1\right]\)
\(f\left(2k\right)=\left(4k^2+1+2k\right)^2+1=\left(4k^2+1\right)^2+4k\left(4k^2+1\right)+4k^2+1\)
\(=\left(4k^2+1\right)\left(4k^2+4k+2\right)=\left(4k^2+1\right)\left[\left(2k+1\right)^2+1\right]\)
\(\Rightarrow\frac{f\left(2k-1\right)}{f\left(2k\right)}=\frac{\left(4k^2+1\right)\left[\left(2k-1\right)^2+1\right]}{\left(4k^2+1\right)\left[\left(2k+1\right)^2+1\right]}=\frac{\left(2k-1\right)^2+1}{\left(2k+1\right)^2+1}\)
\(\Rightarrow\frac{f\left(1\right).f\left(3\right).f\left(5\right)...f\left(2k-1\right)}{f\left(2\right).f\left(4\right).f\left(6\right)...f\left(2k\right)}=\frac{2}{10}.\frac{10}{16}.\frac{16}{50}...\frac{\left(2k-3\right)^2+1}{\left(2k-1\right)^2+1}.\frac{\left(2k-1\right)^2+1}{\left(2k+1\right)^2+1}=\frac{2}{\left(2k+1\right)^2+1}\)
\(\Rightarrow\frac{f\left(1\right)f\left(3\right)...f\left(2017\right)}{f\left(2\right)f\left(4\right)...f\left(2018\right)}=\frac{2}{2019^2+1}=\frac{1}{2038181}\)
