\(CMR\)
\(\frac{5\sqrt{x}-2}{5\sqrt{x}+3}\le\frac{2}{3}\)
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AH
Akai Haruma
Giáo viên
5 tháng 8 2020
Lời giải:
ĐK: $x\geq 0; x\neq 1$
Ta có:
$P(x)=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{15\sqrt{x}-11-(3\sqrt{x}-2)(\sqrt{x}+3)-(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{-5x+13\sqrt{x}-8}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{(8-5\sqrt{x})(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{8-5\sqrt{x}}{\sqrt{x}+3}$
Với $P=\frac{8-5\sqrt{x}}{\sqrt{x}+3}$ thì chưa đủ cơ sở để khẳng định $P(x)\leq \frac{2}{3}$
S
25 tháng 7 2019
1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)
\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)
\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)
Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)
2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)
Áp dụng công thức trên ta được n=2016
3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)
\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)
Thay x=1/3 vào A ta được;
\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)
3 tháng 5 2020
Từ gt => \(\hept{\begin{cases}\left(\frac{1}{\sqrt{2}}-\sqrt{x}\right)\left(\frac{1}{\sqrt{2}}-\sqrt{y}\right)\ge0\Leftrightarrow\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}+\sqrt{2}\sqrt{xy}\left(1\right)\\x\sqrt{x}\le x\cdot\frac{1}{\sqrt{2}};y\sqrt{y}\le y\cdot\frac{1}{\sqrt{2}}\Rightarrow x\sqrt{x}+y\sqrt{y}\le\frac{1}{\sqrt{2}}\left(x+y\right)\left(2\right)\end{cases}}\)
Lại có \(\hept{\begin{cases}\sqrt{xy}\le xy+\frac{1}{4}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{2\sqrt{2}}{3}\sqrt{xy}\le\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)\left(3\right)\\\frac{\sqrt{2}}{3}\sqrt{xy}\le\frac{\sqrt{2}}{6}\left(x+y\right)\left(4\right)\end{cases}}}\)
Từ (1)(2)(3)(4) ta có:\(x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\le\frac{\sqrt{2}}{2}\left(x+y\right)+\frac{\sqrt{2}}{2}+\frac{2\sqrt{2}}{3}\left(xy+\frac{1}{4}\right)+\frac{\sqrt{2}}{6}\left(x+y\right)\)
\(\le\frac{2\sqrt{2}}{3}\left(1+x+y+xy\right)\)
=> \(VT=\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{1+x+y+xy}\le\frac{2\sqrt{2}}{3}\)
Dấu "=" xảy ra <=> x=y=\(\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2019
ĐKXĐ: \(x\ge0;x\ne4;9\)
\(P=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}\right)=\frac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{x-4}\)
\(\frac{1}{P}\le-\frac{5}{2}\Leftrightarrow\frac{1}{P}+\frac{5}{2}\le0\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\)
\(\Rightarrow\sqrt{x}\le\frac{1}{2}\Rightarrow0\le x\le\frac{1}{4}\)