ab+bc+ac=3abc
a,b,c>0
GTNN
\(K=\frac{a^2}{c\left(c^2+a^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\)
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
tính: \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(a^2+ab-c^2-bc\right)}\)
Tính (phân thức)
a)\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\left(\frac{b}{c-a}+\frac{c}{a-b}\right)\)
\(\Rightarrow\frac{a}{b-c}=-\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab-c^2+ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Tương tự:
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)};\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Cộng lại:
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
P/S:Đây nha ! Bài này lớp 7 chắc xem xong key cũng chắc là đang khó hiểu nhỉ ? Đưa giấy bút ra rồi nháp vài cái là hiểu ngay thôi !
Có người nhờ giải ấy @gunny :33
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có :\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
Tương tự : \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(MTC=\left(a-b\right)\left(b-c\right)\left(c-s\right)\left(a+b+c\right)\)
Kí hiệu biểu thức đã cho bởi \(Q\),ta có :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Tính
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Lời giải:
\(\frac{1}{(b-c)(a^2+ac-b^2-bc)}+\frac{1}{(c-a)(b^2+ab-c^2-ac)}+\frac{1}{(a-b)(c^2+bc-a^2-ab)}\)
\(=\frac{1}{(b-c)[(a^2-b^2)+(ac-bc)]}+\frac{1}{(c-a)[(b^2-c^2)+(ab-ac)]}+\frac{1}{(a-b)[(c^2-a^2)+(bc-ab)]}\)
\(=\frac{1}{(b-c)(a-b)(a+b+c)}+\frac{1}{(c-a)(b-c)(b+c+a)}+\frac{1}{(a-b)(c-a)(c+a+b)}\)
\(=\frac{c-a}{(b-c)(a-b)(c-a)(a+b+c)}+\frac{a-b}{(a-b)(c-a)(b-c)(a+b+c)}+\frac{b-c}{(a-b)(c-a)(b-c)(a+b+c)}\)
\(=\frac{c-a+a-b+b-c}{(a-b)(b-c)(c-a)(a+b+c)}=0\)
a, b, c > 0. CMR: \(\left(\frac{ab}{c}\right)^2+\left(\frac{bc}{a}\right)^2+\left(\frac{ac}{b}\right)^2\ge3\left(\frac{ab+bc+ac}{a+b+c}\right)^2\)
\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
Thực hiện phép tính sau:
\(P=\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Xét \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
=> \(P=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(b-c\right)\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}=0\)