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Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath
Học tốt=)
tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2
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
a, x^2 - 2xy + 2y^2 - 2x + 6y + 5 =0
<=> x^2 - 2x(y+1) + y^2 + 2y + 1 + y^2 + 4y + 4 = 0
<=> x^2 - 2x(y+1) + (y+1)^2 + (y+2)^2 =0
<=> (x-y-1)^2 + (y+2)^2 =0
<=> x-y-1 = 0 và y+2 =0
<=> y = -2 => x= -1
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2abc\left(a+b+c\right)}{a^2b^2c^2}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{xyc+yza+zxb}{abc}=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\frac{yza+zxb+xyc}{xyz}=0\)
\(\Rightarrow yza+zxb+xyc=0\)
\(\Rightarrow A=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
\(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)
Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
Thay vào A ta được :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=0\)
Ta có: \(a+b+c=0\)
\(\Rightarrow c=-\left(a+b\right)\)
\(\Rightarrow b=-\left(a+c\right)\)
\(\Rightarrow a=-\left(b+c\right)\)
Thay: \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\) vào \(M\) ta được:
\(M=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(a+c\right)^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}\)
\(=\frac{a+b+c}{-2abc}=0\)