cho x>1,y>0 .Cmr: \(\frac{1}{\left(x-1\right)^3}+\frac{\left(x-1\right)^3}{y^3}+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
cho x>1; y>0, chứng minh \(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
Áp dụng BĐT cô si\(\frac{1}{\left(x-1\right)^3}+1+1\ge\sqrt[3]{\frac{1}{\left(x-1\right)^3}\cdot1\cdot1}=\frac{1}{x-1}\)
\(\Rightarrow\frac{1}{\left(x-1\right)^3}\ge\frac{3}{x-1}-2\left(1\right)\)
\(\left(\frac{x-1}{y}\right)^3+1+1\ge3\sqrt[3]{\left(\frac{x-1}{y}\right)^3\cdot1\cdot1}=\frac{3x-3}{y}\)
\(\Rightarrow\left(\frac{x-1}{y}\right)^3\ge\frac{3x-3}{y}-2\left(2\right)\)
\(\frac{1}{y^3}+1+1\ge\sqrt[3]{\frac{1}{y^3}\cdot1\cdot1}=\frac{3}{y}\Rightarrow\frac{1}{y^3}=\frac{3}{y}-2\left(3\right)\)
Cộng vế theo vế của \(\left(1\right);\left(2\right);\left(3\right)\) ta có:
\(VT\ge\frac{3}{x-1}-6+\frac{3x-3}{y}+\frac{3}{y}\)
\(=\frac{3-6x+6}{x-1}+\frac{3x}{y}\)
\(=3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
Cho x>1, y>0, chứng minh :
\(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
Ta có:
\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\)
\(\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{\left(x-1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)\)
\(=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)
Cho x>1 và y>0. CM
\(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
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Câu hỏi của Phan Thị Hà Vy - Toán lớp 9 - Học toán với OnlineMath
cho x, y >1 cm
\(\frac{1}{\left(x-1\right)^3}+\frac{\left(x-1\right)^3}{y^3}+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
Chứng minh rằng:
Với x>1; y>0 thì \(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
CMR: \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}.\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
Cho x và y là hai số khác 0 và thỏa mãn x+y khác 0. Chứng minh rằng:
\(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
1)tìm các số nguyên x và y thỏa mãn:\(y^2=x^2+x+1\)
2)cho các số thực x và y thỏa mãn \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)=a
tìm giá trị biểu thức \(4\left(x^7+y^7\right)+2\left(x^5+y^5\right)+11\left(x^3+y^3\right)+2016\)
3)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0
cmr \(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(=\frac{1}{x^3y^3}\)
4)cho a,b,c là các số dương.cmr\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)