rút gọn biểu thức
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
Rút gọ biểu thức
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
Rút gọn biểu thức sau :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
Rút gọn phân thức sau :
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho a,b,c là 3 số đôi một khác nhau.Tính giá trị biểu thức
A=\(\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}+\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Chứng minh rằng với a, b, c là các số đôi một khác nhau thì:
\(\frac{a^2\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{c^2\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}=x^2\)
bđt<=>\(S_a\left(a-b\right)^2+S_b\left(b-c\right)^2+S_c\left(c-a\right)^2\ge0\)
with \(S_a=\frac{1}{2\left(a^2+b^2\right)}-\frac{c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(S_b=\frac{1}{2\left(b^2+c^2\right)}-\frac{a}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(S_c=\frac{1}{2\left(c^2+a^2\right)}-\frac{b}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cần cm \(S_a+S_c;S_b+S_c>0\)
lại có:\(S_a+S_c=\frac{1}{2}\left(\frac{1}{a^2+b^2}+\frac{1}{c^2+a^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}\)
\(>\frac{1}{2}\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(c+a\right)^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}>0\)
cmtt=>q.e.d
@Cool Kid:
\(a^3+b^3+c^3+3abc\ge\Sigma ab\sqrt{2\left(a^2+b^2\right)}\)
\(\Leftrightarrow\Sigma\frac{1}{2}\left(a+b-c\right)\left(a-b\right)^2\ge\Sigma\frac{ab\left(a-b\right)^2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)
Hay một BĐT mạnh (và đẹp:v) hơn là:
\(\Leftrightarrow\Sigma\frac{1}{2}\left(a+b-c\right)\left(a-b\right)^2\ge\Sigma\frac{ab\left(a-b\right)^2}{2\left(a+b\right)}\)
Ta cần chứng minh: \(VT-VP=\Sigma\frac{\left(a+b-c\right)^2\left(a-b\right)^2}{2\left(a+b\right)}-\frac{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
Giả sử \(a\ge c\ge b\) và đặt \(a=b+u+v,c=b+v\)
Bất đẳng thức này đúng theo Cauchy-Schwawrz:
\(VT-VP\ge\frac{4\left(c+a-b\right)^2\left(c-a\right)^2}{4\left(a+b+c\right)}-\frac{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
Last inequality is: https://imgur.com/tRsHOfr (mình không gửi ảnh được nên gửi link vậy!)
Done!