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 !!!
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)
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)
Tìm GTNN của biểu thức : \(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\) biết \(x>0\).
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 :)
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
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 \(x=\frac{1}{x}\Leftrightarrow x=1\)
Đúng 0
Bình luận (0)
1. Gỉai các phương trình sau
\(a,\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(b,\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(c,\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
HELP ME!
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow27x-2x-4x-27+2=0\)
\(\Leftrightarrow21x=25\)
\(\Leftrightarrow x=\frac{25}{21}\)
Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !
\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)
\(\Leftrightarrow-20x-12=56\)
\(\Leftrightarrow-20x=68\)
\(\Leftrightarrow x=-\frac{17}{5}\)
Tự check lại nhá
\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Ta dễ thấy \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}< 0\) nên
\(x+95=0\Leftrightarrow x=-95\)
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
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)