Các câu hỏi tương tự
rút gọn phân thức\(\frac{a^2\cdot\left(b-c\right)+b^2\cdot\left(c-a\right)+c^2\cdot\left(a-b\right)}{a^4\cdot\left(b^2-c^2\right)+b^4\cdot\left(c^2-a^2\right)+c^4\cdot\left(a^2-b^2\right)}\)
Rút gọn các phân thức sau
a) \(A=\frac{a^2\cdot\left(b-c\right)+b^2\cdot\left(c-a\right)+c^2\cdot\left(a-b\right)}{a\cdot b^2-a\cdot c^2-b^3+b\cdot c^2}\)
b) \(B=\frac{x^3+y^3+z^3-3\cdot x\cdot y\cdot z}{\left(x+y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Cho a,b,c đôi một khác nhau
Tính P=\(\frac{a^2}{\left(a-b\right)\cdot\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\cdot\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\cdot\left(c-a\right)}\)
rút gọn bt
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}\)
chứng minh\(\frac{a\cdot\left(b+c\right)}{a^2+2bc}+\frac{b\cdot\left(a+c\right)}{b^2+2ac}+\frac{c\cdot\left(a+b\right)}{c^2+2ab}< =2\)2 với a,b,c là độ dài 3 cạnh tam giác
Tính \(y=\frac{a\cdot b}{\left(b-c\right)\cdot\left(c-a\right)}+\frac{b\cdot c}{\left(c-a\right)\cdot\left(a-b\right)}+\frac{a\cdot c}{\left(a-b\right)\cdot\left(b-c\right)}\)
Cho \(\hept{\begin{cases}a\cdot\left(b^{2+c^2}\right)+b\cdot\left(b^2+c^2\right)+c\left(a^2+b^2\right)+2abc=0\\a^{3+}b^3+c^3=1\end{cases}Tính}A=\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}\left(a,b,c#0\right)\)
Mạnh mẽ hơn Nesbitt?Với a, b, c là các số thực sao cho: a+b+c0,text{ }ab+bc+ca0,text{ }left(a+bright)left(b+cright)left(c+aright)0 thì:frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}-frac{3}{2}geleft(Sigma abright)left(Sigmafrac{1}{left(a+bright)^2}right)-frac{9}{4}Chứng minh: 4left(a+b+cright)left(a+bright)^2left(b+cright)^2left(c+aright)^2cdotleft(text{VT}-text{VP}right)left(a+bright)left(b+cright)left(c+aright)left[Sigmaleft(ab+bc-2caright)^2+left(ab+bc+caright)Sigmaleft(a-bright)^2right]+left(a+b+cri...
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Mạnh mẽ hơn Nesbitt?
Với a, b, c là các số thực sao cho: \(a+b+c>0,\text{ }ab+bc+ca>0,\text{ }\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\) thì:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\ge\left(\Sigma ab\right)\left(\Sigma\frac{1}{\left(a+b\right)^2}\right)-\frac{9}{4}\)
Chứng minh: \(4\left(a+b+c\right)\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\cdot\left(\text{VT}-\text{VP}\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\left[\Sigma\left(ab+bc-2ca\right)^2+\left(ab+bc+ca\right)\Sigma\left(a-b\right)^2\right]\)
\(+\left(a+b+c\right)\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\ge0\)
\(\left(1+\frac{b^2+c^2-a^2}{2bc}\right)\cdot\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}\cdot\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)