142.
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)
\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\dfrac{\left(x+y\right)\left(xy+yz+zx+z^2\right)}{xyz\left(x+y+z\right)}=0\)
\(\Rightarrow\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+y+z=z\\x+y+z=x\\x+y+z=y\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=z\\a=x\\a=y\end{matrix}\right.\)
143.
Biểu thức xác định khi \(xyz\ne0\)
Giả sử \(x+y+z\) và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) đồng thời bằng 0
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\)
\(\Rightarrow xy+yz+zx=0\) (1)
Lại có:
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\) (2)
\(\left(1\right);\left(2\right)\Rightarrow x^2+y^2+z^2=0\)
\(\Rightarrow x=y=z=0\) (ko thỏa mãn ĐKXĐ)
Vậy điều giả sử là sai hay \(x+y+z\) và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) ko thể đồng thời bằng 0
144.
\(\left\{{}\begin{matrix}2a=by+cz\left(1\right)\\2b=ax+cz\left(2\right)\\2c=ax+by\left(3\right)\end{matrix}\right.\)
Cộng vế (1), (2) và (3):
\(\Rightarrow a+b+c=ax+by+cz\) (4)
Lần lượt trừ vế (4) cho (1), (2) và (3):
\(\Rightarrow\left\{{}\begin{matrix}ax=b+c-a\\by=a+c-b\\cz=a+b-c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{b+c-a}{a}\\y=\dfrac{a+c-b}{b}\\z=\dfrac{a+b-c}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=\dfrac{a+b+c}{a}\\y+2=\dfrac{a+b+c}{b}\\z+2=\dfrac{a+b+c}{c}\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\)
145.
a.
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc.c}{ac+abc.c+abc}\)
\(=\dfrac{a}{a\left(bc+b+1\right)}+\dfrac{b}{bc+b+1}+\dfrac{ac.bc}{ac\left(bc+b+1\right)}\)
\(=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(=\dfrac{bc+b+1}{bc+b+1}=1\)
b.
\(N=\dfrac{a}{ab+a+1}+\dfrac{b.abc}{bc+b.abc+abc}+\dfrac{c}{ac+c+abc}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab.bc}{bc\left(ab+a+1\right)}+\dfrac{c}{c\left(ab+a+1\right)}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{ab+a+1}+\dfrac{1}{ab+a+1}=1\)
146.
\(\dfrac{a}{c}=\dfrac{a-b}{b-c}\Rightarrow\dfrac{1}{a}=\dfrac{b-c}{c\left(a-b\right)}\)
Đồng thời \(\dfrac{1}{c}=\dfrac{a-b}{a\left(b-c\right)}\)
\(\dfrac{a}{c}=\dfrac{a-b}{b-c}\Rightarrow c\left(a-b\right)=a\left(b-c\right)\) (1)
Do đó:
\(\dfrac{1}{a}+\dfrac{1}{a-b}=\dfrac{b-c}{c\left(a-b\right)}+\dfrac{1}{a-b}=\dfrac{b-c+c}{c\left(a-b\right)}=\dfrac{b}{c\left(a-b\right)}\) (2)
\(\dfrac{1}{b-c}-\dfrac{1}{c}=\dfrac{1}{b-c}-\dfrac{a-b}{a\left(b-c\right)}=\dfrac{a-\left(a-b\right)}{a\left(b-c\right)}=\dfrac{b}{a\left(b-c\right)}\) (3)
(1);(2);(3) \(\Rightarrow\dfrac{b}{c\left(a-b\right)}=\dfrac{b}{a\left(b-c\right)}\)
Hay \(\dfrac{1}{a}+\dfrac{1}{a-b}=\dfrac{1}{b-c}-\dfrac{1}{c}\)
147.
a.
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(-c\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow\dfrac{a^3+b^3+c^3}{abc}=3\)
\(\Rightarrow\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=3\)
b.
\(B=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\dfrac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\dfrac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\dfrac{a^2}{\left(a-b\right).\left(-c\right)-c^2}+\dfrac{b^2}{\left(b-c\right).\left(-a\right)-a^2}+\dfrac{c^2}{\left(c-a\right)\left(-b\right)-b^2}\)
\(=\dfrac{-a^2}{c\left(a+c-b\right)}-\dfrac{b^2}{a\left(a+b-c\right)}-\dfrac{c^2}{b\left(b+c-a\right)}\)
\(=\dfrac{-a^2}{c\left(-b-b\right)}-\dfrac{b^2}{a\left(-c-c\right)}-\dfrac{c^2}{b\left(-a-a\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\right)=\dfrac{1}{2}.3=\dfrac{3}{2}\)
148.
\(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}=-\dfrac{a-b}{a+b}+\dfrac{b+\left(a+b\right)}{a}+\dfrac{-\left(a+b\right)-a}{b}\)
\(=\dfrac{a+2b}{a}-\dfrac{2a+b}{b}-\dfrac{a-b}{a+b}\)
\(=1+\dfrac{2b}{a}-\dfrac{2a}{b}-1-\dfrac{a-b}{a+b}=\dfrac{2\left(b^2-a^2\right)}{ab}+\dfrac{b-a}{a+b}\)
\(=\left(b-a\right)\left[\dfrac{2\left(a+b\right)}{ab}+\dfrac{1}{a+b}\right]=\dfrac{\left(b-a\right)\left(2a^2+5ab+2b^2\right)}{ab\left(a+b\right)}\)
\(=\dfrac{\left(b-a\right)\left(a+2b\right)\left(2a+b\right)}{ab\left(a+b\right)}\) (1)
\(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}=\dfrac{-\left(a+b\right)}{a-b}+\dfrac{a}{b+\left(a+b\right)}+\dfrac{b}{-\left(a+b\right)-a}\)
\(=\dfrac{a+b}{b-a}+\dfrac{a}{a+2b}-\dfrac{b}{2a+b}=\dfrac{a+b}{b-a}+\dfrac{2a^2-2b^2}{\left(a+2b\right)\left(2a+b\right)}\)
\(=\dfrac{a+b}{b-a}+\dfrac{2\left(a+b\right)\left(a-b\right)}{\left(a+2b\right)\left(2a+b\right)}=\left(a+b\right)\left(\dfrac{1}{b-a}+\dfrac{2\left(a-b\right)}{\left(a+2b\right)\left(2a+b\right)}\right)\)
\(=\left(a+b\right).\dfrac{9ab}{\left(b-a\right)\left(a+2b\right)\left(2a+b\right)}\) (2)
(1);(2) \(\Rightarrow A=\dfrac{\left(b-a\right)\left(a+2b\right)\left(2a+b\right)}{ab\left(a+b\right)}.\dfrac{9ab\left(a+b\right)}{\left(b-a\right)\left(a+2b\right)\left(2a+b\right)}=9\)
149.
\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right)\)
\(\Leftrightarrow a^2b-a^3bc-b^2c+ab^2c^2=ab^2-ab^3c-a^2c+a^2bc^2\)
\(\Leftrightarrow\left(a^2b-ab^2\right)+a^2c-b^2c-\left(a^3bc-ab^3c\right)+ab^2c^2-a^2bc^2=0\)
\(\Leftrightarrow ab\left(a-b\right)+\left(ac+bc\right)\left(a-b\right)-abc\left(a-b\right)\left(a+b\right)-abc^2\left(a-b\right)=0\)
\(\Leftrightarrow ab+ac+bc-abc\left(a+b\right)-abc^2=0\)
\(\Leftrightarrow ab+bc+ca-abc\left(a+b+c\right)=0\)
\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}=a+b+c\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)