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1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
\(PT\Leftrightarrow\left(\dfrac{x-b-c}{a}-1\right)+\left(\dfrac{x-a-c}{b}-1\right)+\left(\dfrac{x-a-b}{c}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-a-b-c}{a}+\dfrac{x-a-b-c}{b}+\dfrac{x-a-b-c}{c}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\) nên \(x-a-b-c=0\Rightarrow x=a+b+c\)
Vậy nghiệm của PT là \(x=a+b+c\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
=\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)\)
mà a+b+c=0
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{0}{abc}\right)=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)
b)Ta có: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)
\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)
\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)
\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)
\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)
cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay kết quả vừa chúng minh vào đề bài ta đc
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
chúc bạn học tốt ^ ^
Lời giải:
Vì \(2a-b=5\Rightarrow b=2a-5\Rightarrow 2b=4a-10\)
\(\Rightarrow 7a-2b=7a-(4a-10)=3a+10\)
\(\Rightarrow \frac{7a-2b}{3a+10}=\frac{3a+10}{3a+10}=1\)
Lại có:
\(2a-b=5\Rightarrow 2a=b+5\Rightarrow 4a=2b+10\)
\(\Rightarrow 7b-4a=7b-(2b+10)=5b-10\)
\(\Rightarrow \frac{7b-4a}{15b-30}=\frac{5b-10}{15b-30}=\frac{5b-10}{3(5b-10)}=\frac{1}{3}\)
Vậy: \(A=1-\frac{1}{3}=\frac{2}{3}\)
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b}\Leftrightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2=ab\)
\(\Rightarrow\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{a+b}{ab}\Rightarrow\dfrac{a+b}{ab}=\dfrac{1}{a+b}\)
\(\Leftrightarrow ab=\left(a+b\right)^2\)
Vậy, \(\dfrac{b}{a}+\dfrac{a}{b}=\dfrac{a^2+b^2}{ab}=\dfrac{a^2+2ab+b^2-2ab}{ab}=\dfrac{\left(a+b\right)^2-2ab}{ab}=\dfrac{ab-2ab}{ab}=\dfrac{-ab}{ab}=-1\)