\(a+b+c=3abc\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3\left(chia:abc\right);\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=1\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-5\)
Cho \(\left\{{}\begin{matrix}a+b+c=abc\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\\a,b,c\ne0\end{matrix}\right.\)
Tính \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Cho a, b, c ≠ 0 thoả mãn \(\left\{{}\begin{matrix}a+b+c=2021\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2021}\end{matrix}\right.\) . Chứng minh: \(\frac{1}{a^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{a^{2021}+b^{2021}+c^{2021}}\)
ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
Cho \(\left\{{}\begin{matrix}ax+by+cz=0\\a+b+c=\frac{1}{2019}\end{matrix}\right.\) . Tính giá trị của \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}\)
Cho x, y, z khác 0 thỏa mãn: \(\left\{{}\begin{matrix}x+y+z=\frac{1}{2}\\\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\end{matrix}\right.\)
Tính: \(P=\left(y^{2009}+z^{2009}\right)\left(z^{2011}+x^{2011}\right)\left(x^{2013}+y^{2013}\right)\)
Giúp hộ mik ạ!!!
a, CMR: 9x2y2+ y2- 6xy - 2y +2≥0
b, cho ba số thuộc số âm x, y, z
thỏa mãn\(\left\{{}\begin{matrix}xyz=1\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}< x+y+z\end{matrix}\right.\)
CMR: Có đúng trong ba số x,y, z lớn hơn 1
1,cho a,b,c là số thực dương thỏa mãn
\(\frac{a^2}{\left(b-c\right)^2}+\frac{b^2}{\left(c-a\right)^2}+\frac{c^2}{\left(a-b\right)^2}=3\)
và \(\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}+\frac{1}{\left(a-b\right)^2}=1\)
Tính
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\)
xác định các số hữu tỉ a,b,c,d sao cho:
a,\(\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{a}{x\left(x+1\right)}+\frac{b}{\left(x+1\right)\left(x+2\right)}\)
b,\(\frac{x^3}{x^4-1}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}\)
c,\(\frac{2x^2-x+1}{\left(x+1\right)\left(x-2\right)^2}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{x-2}\)
Cho 3 số thực dương a, b, c.
Chứng minh rằng: \(\frac{b}{a\left(a+b\right)}+\frac{c}{b\left(b+c\right)}+\frac{a}{c\left(c+a\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
