120.
\(A=\dfrac{x+3a}{2-x}+\dfrac{x-3a}{2+x}-\dfrac{2a}{4-x^2}+a\)
\(=\dfrac{\left(x+3a\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{\left(x-3a\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}-\dfrac{2a}{\left(2-x\right)\left(2+x\right)}+a\)
\(=\dfrac{x^2+\left(3a+2\right)x+6a-x^2+\left(3a+2\right)x-6a-2a}{\left(2-x\right)\left(2+x\right)}+a\)
\(=\dfrac{2\left(3a+2\right)x-2a}{4-x^2}+a\)
Do \(x=\dfrac{a}{3a+2}\Rightarrow2\left(3a+2\right)x=2a\)
\(\Rightarrow A=\dfrac{2a-2a}{4-x^2}+a=a\)
121.
Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\)
\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)
Ta có:
\(A=\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(=\dfrac{2\left(xy+yz+zx\right)}{xyz}+\dfrac{x^2+y^2+z^2}{xyz}\)
\(=\dfrac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{xyz}\)
\(=\dfrac{\left(x+y+z\right)^2}{xyz}=\dfrac{0^2}{xyz}=0\)
122.
\(\dfrac{a+b-c}{ab}-\dfrac{b+c-a}{bc}-\dfrac{a+c-b}{ac}=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{c}{ab}\right)-\left(\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right)-\left(\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{b}{ac}\right)=0\)
\(\Leftrightarrow\dfrac{a}{bc}+\dfrac{b}{ca}-\dfrac{c}{ab}-\dfrac{2}{c}=0\)
\(\Leftrightarrow\dfrac{a^2+b^2-c^2-2ab}{abc}=0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2-c^2}{abc}=0\)
\(\Leftrightarrow\dfrac{\left(a-b+c\right)\left(a-b-c\right)}{abc}=0\)
\(\Rightarrow\left[{}\begin{matrix}a+c-b=0\\b+c-a=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{b+c-a}{bc}=0\\\dfrac{a+c-b}{ac}=0\end{matrix}\right.\)
Hay có ít nhất 1 phân thức đã cho bằng 0
124.
\(B=\left(ab+bc+ca\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
\(\Rightarrow B=abc\left[\dfrac{\left(ab+bc+ca\right)}{abc}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]\)
\(=abc\left[\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]\)
\(=abc\left[\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right]\)
\(=abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}-\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right)\)
\(=abc\left(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)=\dfrac{2abc\left(a+b+c\right)}{abc}\)
\(=2\left(a+b+c\right)\)
125.
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\)
\(\Rightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}ab=-\left(bc+ca\right)\\bc=-\left(ab+ca\right)\\ca=-\left(ab+bc\right)\end{matrix}\right.\)
a.
\(M=\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)
\(=\dfrac{1}{a^2+bc-\left(ab+ca\right)}+\dfrac{1}{b^2+ca-\left(ab+bc\right)}+\dfrac{1}{c^2+ab-\left(bc+ca\right)}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{a\left(b-c\right)-c\left(b-c\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(b-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
125b.
Áp dụng kết quả mẫu số câu a:
\(N=\dfrac{bc}{a^2+2bc}+\dfrac{ca}{b^2+2ca}+\dfrac{ab}{c^2+2ab}\)
\(=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}-\dfrac{ca}{\left(a-b\right)\left(b-c\right)}+\dfrac{ab}{\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{bc\left(b-c\right)-ca\left(a-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2c-bc^2-a^2c+ac^2+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{c^2\left(a-b\right)-c\left(a^2-b^2\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{c^2\left(a-b\right)-\left(ac+bc\right)\left(a-b\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(b\left(a-c\right)-c\left(a-c\right)\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
125c.
\(P=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
\(=\left(\dfrac{a^2}{a^2+2bc}-1\right)+\left(\dfrac{b^2}{b^2+2ca}-1\right)+\left(\dfrac{c^2}{c^2+2ab}-1\right)+3\)
\(=-2\left(\dfrac{bc}{a^2+2bc}+\dfrac{ca}{b^2+2ca}+\dfrac{ab}{c^2+2ab}\right)+3\)
\(=-2.1+3=1\)
126.
Nếu \(a+b+c\ne0\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c=3c\\a+b+c=3a\\a+b+c=3b\end{matrix}\right.\)
\(\Rightarrow3a=3b=3c\)
\(\Rightarrow a=b=c\) (ko thỏa mãn điều kiện a;b;c đôi một khác nhau)
\(\Rightarrow a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+b=-b\end{matrix}\right.\)
\(\Rightarrow M=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)=\dfrac{-abc}{abc}=-1\)