cho các số abc thỏa mãn 1/a+b=2/b+c=3/c+a tính q=a+2021b+c/a+2022b+c
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Áp dụng t/c dtsbn:
\(\dfrac{1}{a+b}=\dfrac{2}{b+c}=\dfrac{3}{c+a}=\dfrac{1+2+3}{2\left(a+b+c\right)}=\dfrac{6}{2\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)
\(\Rightarrow\left\{{}\begin{matrix}3a+3b=a+b+c\\3b+3c=2a+2b+2c\\3a+3c=3a+3b+3c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=2a\\b=0\end{matrix}\right.\)
\(Q=\dfrac{a+2021b+c}{a+2022b+c}=\dfrac{a+2a}{a+2a}=1\)
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\(\frac{1}{a+b}=\frac{2}{b+c}=\frac{3}{c+a}=\frac{1+2+3}{2\left(a+b+c\right)}=\frac{3}{a+b+c}.\)
\(\Rightarrow\frac{3}{c+a}=\frac{3}{a+b+c}\Rightarrow c+a=a+b+c\Rightarrow b=0\)
\(\Rightarrow Q=\frac{a+2021b+c}{a+2022b+c}=\frac{a+c}{a+c}=1\)
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Ta có: aba+b=bcb+c=aca+c⇒a+bab=b+cbc=a+cacaba+b=bcb+c=aca+c⇒a+bab=b+cbc=a+cac
⇒aab+bab=bbc+cbc=aac+cac⇒aab+bab=bbc+cbc=aac+cac
⇒1b+1a=1c+1b=1c+1a⇒1b+1a=1c+1b=1c+1a
⇒1a=1b=1c⇒a=b=c⇒1a=1b=1c⇒a=b=c
⇒M=ab+bc+caa2+b2+c2= a2+b2+c2a2+b2+c2=1
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\(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+c}\Rightarrow\frac{1}{a+b}=\frac{1}{b+c}\Rightarrow a+b=b+c\)
\(\Rightarrow a=c\left(1\right)\)
\(\frac{1}{b+c}=\frac{1}{c+c}\Rightarrow b+c=c+c\Rightarrow c=b\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=b=c\)
\(Q=\frac{a+2021b+c}{a+2022b+c}=\frac{a+2021a+a}{a+2022a+a}\)
\(Q=\frac{a.\left(1+2021+1\right)}{a.\left(1+2022+1\right)}=\frac{2023}{2024}\)
Vậy, \(Q=\frac{2023}{2024}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)
\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{-abc}{abc}=-1\)
Với \(a+b+c\ne0\)
\(\dfrac{a+b-2021c}{c}=\dfrac{b+c-2021a}{a}=\dfrac{c+a-2021b}{b}=\dfrac{-2019\left(a+b+c\right)}{a+b+c}=-2019\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-2021c=-2019c\\b+c-2021a=-2019a\\c+a-2021b=-2019b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)
Với a+b+c=0⇔⎧⎪⎨⎪⎩b+c=−ac+a=−ba+b=−ca+b+c=0⇔{b+c=−ac+a=−ba+b=−c
B=a+ba⋅a+cc⋅b+cb=−abcabc=−1B=a+ba⋅a+cc⋅b+cb=−abcabc=−1
Với a+b+c≠0a+b+c≠0
a+b−2021cc=b+c−2021aa=c+a−2021bb=−2019(a+b+c)a+b+c=−2019⇔⎧⎪⎨⎪⎩a+b−2021c=−2019cb+c−2021a=−2019ac+a−2021b=−2019b⇔⎧⎪⎨⎪⎩a+b=2cb+c=2ac+a=2b
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Lời giải:
$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$
$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$
$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$
$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$
$\Rightarrow a+b+c=0$ hoặc $a=b=c$
Nếu $a+b+c=0$ thì:
$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:
$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\)
\(=\sqrt{2a\left(a+b+c\right)+\dfrac{b^2-2bc+c^2}{2}}\)
\(=\sqrt{\dfrac{4a^2+b^2+c^2+4ab+4ac-2bc}{2}}\)
\(=\sqrt{\dfrac{\left(2a+b+c\right)^2-4bc}{2}}\)
\(\le\sqrt{\dfrac{\left(2a+b+c\right)^2}{2}}\)
\(=\dfrac{2a+b+c}{\sqrt{2}}\).
Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)
\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)
ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,b\)là hai nghiệm phân biệt của phương trình: \(x^2-2021x-c=0\).
Theo Viet:
\(\hept{\begin{cases}a+b=2021\\ab=-c\end{cases}}\)
\(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{2021}{-c}\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{2021}{c}=0\)