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\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\Leftrightarrow a^2+b^2+2ab=c^2\Leftrightarrow a^2+b^2-c^2=-2ab\)
tương tự ta có: b2+c2-a2=-2bc ; a2+c2-b2=-2ac
Do đó \(P=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Leftrightarrow bc+ca+ab=0\)
\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)
Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
Ta có:\(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Rightarrow a^2+b^2+2ab=c^2\Rightarrow a^2+b^2-c^2=-2ab\)
Tươmg tự ta cũng có:\(b^2+c^2-a^2=-2bc\) và \(c^2+a^2-b^2=-2ca\)
\(\Rightarrow P=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ca}=-\frac{1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0\)
a+b+c=0 => a= -(b+c) TƯƠNG TỰ
b= -(a+c) ; c= -(b+a)
ta co P= \(\frac{1}{\left(b+c\right)^2+\left(b^2-c^2\right)}+\frac{1}{\left(a+c\right)^2+\left(a^2-c^2\right)}+\frac{1}{\left(b+a\right)^2+\left(b^2-a^2\right)}\)
=> P= \(\frac{1}{2c\left(b+c\right)}+\frac{1}{2b\left(a+c\right)}+\frac{1}{2a\left(b+c\right)}\)
thay b+c=-a; a+c=-b ; a+b=-c (như trên )
=> P= \(\frac{1}{-2ac}+\frac{1}{-2ab}+\frac{1}{-2bc}\)
QUY ĐONG CAC MAU THUC TA CO
P= \(\frac{a+b+c}{-2abc}\)
a+b+c=0 => P=0
Câu 1 :
\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2-x^2}+\frac{1}{y^2+2xy+x^2}\right)\)
a) ĐKXĐ : \(x\ne\pm y\)
b) Ta có : \(A=\frac{4xy}{\left(y-x\right)\left(x+y\right)}:\left(\frac{1}{\left(y-x\right)\left(x+y\right)}+\frac{1}{\left(x+y\right)^2}\right)\)
\(=\frac{4xy}{\left(y-x\right)\left(x+y\right)}:\left(\frac{x+y+y-x}{\left(x+y\right)^2\left(y-x\right)}\right)\)
\(=\frac{4xy}{\left(y-x\right)\left(x+y\right)}\cdot\frac{\left(x+y\right)^2\left(y-x\right)}{2y}\)
\(=2x\left(x+y\right)\)
Vậy : \(A=2x\left(x+y\right)\) với \(x\ne\pm y\)
b/ \(\Leftrightarrow A=\frac{4xy}{y^2-x^2}-\left(y^2-x^2\right)+\frac{4xy}{\left(y-x\right)\left(x+y\right)}.\left(x+y\right)^2\)
\(\Leftrightarrow A=4xy+\frac{4x^2y+4xy^2}{y-x}\)
\(\Leftrightarrow A=4xy.\left(1+\frac{x+y}{y-x}\right)\)
\(\Leftrightarrow A=\frac{8xy^2}{y-x}\)
Tham khao =))
Câu hỏi của Vu Quang Huy - Toán lớp 8 - Học toán với OnlineMath