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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:
$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$
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
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
\(\Leftrightarrow\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}=\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)
Ta có: \(\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)(cmt)
nên \(\dfrac{2020a^2+2021b^2}{2020a^2-2021b^2}=\dfrac{2020c^2+2021d^2}{2020c^2-2021d^2}\)(đpcm)
từ giả thiết \(\Rightarrow\frac{b}{a}=\frac{d}{c}\)
ta có
\(\frac{2021a-b}{a}=\frac{2021a}{a}-\frac{b}{a}=2021-\frac{b}{a}=2021-\frac{d}{c}=\frac{2021c-d}{c}\)
vậy ta có dpcm
a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)
\(\left(b-1\right)^{2024}>=0\forall b\)
Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)
Thay a=-1 và b=1 vào P, ta được:
\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)
a^2=bc
=>a*a=b/c
=>a/b=c/a=k
=>a=bk; c=ak
\(\dfrac{2022a+2021b}{2023a-2024b}=\dfrac{2022\cdot bk+2021b}{2023\cdot bk-2024b}=\dfrac{2022k+2021}{2023k-2024}\)
\(\dfrac{2022c+2021a}{2023c-2024a}=\dfrac{2022ak+2021a}{2023ak-2024a}=\dfrac{2022k+2021}{2023k-2024}\)
=>\(\dfrac{2022a+2021b}{2023a-2024b}=\dfrac{2022c+2021a}{2023c-2024a}\)