Tìm min P với x>0 \(p=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
Những câu hỏi liên quan
Cho x>0 Tìm GTNN
\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
Ta có :
\(P=\frac{\left(x+\frac{1}{x}^6\right)-\left(x^6+\frac{1}{x}^6\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x}^3\right)\)
\(=3\left(x+\frac{1}{x}\right)\ge6\left(x>0\right)\)
\(\Rightarrow Pmin=6\Leftrightarrow x=1\)
Đúng 0
Bình luận (0)
Dleft ( frac{1}{3sqrt{x}-6} +frac{1}{x-2sqrt{x}}right )left ( frac{1}{6} +frac{1}{2sqrt{x}}right ) Dleft ( frac{1}{3left ( sqrt{x}-2 right )} +frac{1}{sqrt{x}left ( sqrt{x}-2 right )}right ).frac{sqrt{x}+3}{6sqrt{x}} Dfrac{sqrt{x}+3}{3sqrt{x}left ( sqrt{x}-2 right )}.frac{sqrt{x}+3}{6sqrt{x}} Dfrac{left ( sqrt{x}+3 right )^{2}}{18xleft ( sqrt{x}-2 right )} Dfrac{x+6sqrt{x}+9}{18xsqrt{x}-36x}
A/ Đúng
B/ Sai
Đọc tiếp
\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
Cho x >0 . tìm GTNN của :
\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
help me !!!
\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left[\left(x^3\right)^2+2x^3\cdot\frac{1}{x^3}+\left(\frac{1}{x^3}\right)^2\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\frac{\left[\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right]\left[\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)
\(=\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\ge\left(2\sqrt{x\cdot\frac{1}{x}}\right)^3+2\sqrt{x^3\cdot\frac{1}{x^3}}=8+2=10\)
Dấu "=" khi x = 1
Đúng 0
Bình luận (0)
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
Tìm GTNN
\(A=\frac{x^2+y^2}{\left(x+y\right)^2}\)
\(B=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}\)\(x>0\)
\(C=\frac{a^2}{x}+\frac{b^2}{y}\)(a và b là hằng số dương đã cho)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
B=\(\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
Rút gọn biểu thức \(B=\left(\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2\right):\left(\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}\right)\)