Cho M=\(\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
a) Rút gọn M
b) Cho x > 0.Tìm GTNN của M
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7 tháng 8 2019
\(M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)+x^3+\frac{1}{x^3}}\)
\(M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\frac{2x^6+3x^4+3x^2+2}{x^3}}\)
\(M=\frac{\left[\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2\right]x^3}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{x^3\left(6x^4+15x^2+\frac{15}{x^2}+\frac{6}{x^4}+18\right)}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{\frac{6x^8+15x^6+18x^4+15x^2+6}{x^4}.x^3}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{\frac{6x^8+15x^6+18x^4+15x^2+6}{x}}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{6x^8+15x^6+18x^4+15x^2+6}{x\left(2x^6+3x^4+3x^2+2\right)}\)
\(M=\frac{3\left(x^2+1\right)^2\left(2x^4+x^2+2\right)}{x\left(x^2+1\right)\left(2x^4+x^2+2\right)}\)
\(M=\frac{3\left(x^3+1\right)}{x}\)
15 tháng 1 2017
a/ Đặt: \(x+\frac{1}{x}=a\)
Ta có: \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)
\(x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2=\left(\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right)^2-2\)
\(=\left(a^3-3a\right)^2-2\)
\(\Rightarrow M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^6-\left(a^3-3a\right)^2+2-2}{a^3+a^3-3a}\)
\(=\frac{\left(a^3+a^3-3a\right)\left(a^3-a^3+3a\right)}{\left(a^3+a^3-3a\right)}=3a\)
\(=3.\left(x+\frac{1}{x}\right)=\frac{3x^2+3}{x}\)
b/ \(\frac{3x^2+3}{x}=3x+\frac{3}{x}\ge2.3=6\)
Đấu = xảy ra khi \(x=\frac{1}{x}\Leftrightarrow x=1\)
LD
11 tháng 6 2019
\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left[\left(x^3\right)^2+2x^3\cdot\frac{1}{x^3}+\left(\frac{1}{x^3}\right)^2\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left[\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right]\left[\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\ge\left(2\sqrt{x\cdot\frac{1}{x}}\right)^3+2\sqrt{x^3\cdot\frac{1}{x^3}}=8+2=10\)
Dấu "=" khi x = 1
24 tháng 11 2018
Ta có :
\(P=\frac{\left(x+\frac{1}{x}^6\right)-\left(x^6+\frac{1}{x}^6\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x}^3\right)\)
\(=3\left(x+\frac{1}{x}\right)\ge6\left(x>0\right)\)
\(\Rightarrow Pmin=6\Leftrightarrow x=1\)
20 tháng 6 2017
Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)^3=a\\x^3+\frac{1}{x^3}=b\end{cases}}\)
Ta có
\(A=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+2+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^2-b^2}{a+b}=a-b\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}-\left(x^3+\frac{1}{x^3}\right)=\frac{3x^2+3}{x}\)
1 tháng 1 2017
Ta có : \(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}-2\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3\left(x+\frac{1}{x}\right)\ge6\) \(\left(x>0\right)\).
Vậy \(P_{Min}=6\) khi \(x=1.\)
Happy New year :)
