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Ta có \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2020}}{a_{2021}}=\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\)(dãy tỉ só bằng nhau)
=> \(\frac{a_1}{a_2}=\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\)
<=> \(\left(\frac{a_1}{a_2}\right)^{2020}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}...\frac{a_{2020}}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)
Khi đó M = 4(a - b)(b - c) - (c - a)2
= 4(2020k - 2021k)(2021k - 2022k) - (2022k - 2020k)2
= 4(-k)(-k) - (2k)2
= 4k2 - 4k2 = 0
Vậy M = 0
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\)( \(k\ne0\))
\(\Rightarrow a=2020k\); \(b=2021k\); \(c=2022k\)
Thay a, b, c vào biểu thức M ta có:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
\(=4\left(2020k-2021k\right)\left(2021k-2022k\right)-\left(2022k-2020k\right)^2\)
\(=4.\left(-k\right).\left(-k\right)-\left(2k\right)^2=4k^2-4k^2=0\)
Vậy \(M=0\)
a) Ta có \(\left(x-2\right)^2\ge0\forall x\)
=> Min A = 0
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy Min A = 0 <=> x = 2
b) Ta có \(\left(2x+1\right)^4\ge0\forall x\Rightarrow\left(2x+1\right)^4-98\ge-98\)
=> Min B = -98
Dấu "=" xảy ra <=> 2x + 1= 0 <=> x = -0,5
Vậy Min B = -98 <=> x = -0,5
c) Ta có C = |x - 10| + |x - 11|
= |x - 10| + |11 - x| \(\ge\left|x-10+11-x\right|=\left|1\right|=1\)
=> Min C = 1
Dấu "=" xảy ra <=> \(\left(x-10\right)\left(11-x\right)\ge0\)
TH1 : \(\hept{\begin{cases}x-10\ge0\\11-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge10\\x\le11\end{cases}}\Leftrightarrow10\le x\le11\)
TH2 : \(\hept{\begin{cases}x-10\le0\\11-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le10\\x\ge11\end{cases}}\Leftrightarrow x\in\varnothing\)
Vậy Min C = 1 <=> \(10\le x\le11\)
\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)
ta có
\(f\left(-2021\right)=f\left(2021\cdot\left(-1\right)\right)=f\left(2021-1\right)=f\left(2020\right)\)
vậy \(f\left(2020\right)=f\left(-2021\right)=2020\)
Sửa đề chứng minh : 4(a - b)(b - c) = (c - a)2
Đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)
Khi đó 4(a - b)(b - c) = 4(2020k - 202k)(2021k - 2022k) = 4(-k)(-k) = 4k2 (1)
Lại có (c- a)2 = (2022k - 2020k)2 = (2k)2 = 4k2 (2)
Từ (1)(2) => 4(a - b)(b - c) = (c - a)2 (đpcm)
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\)
\(=3+Q\)
Suy ra \(3+Q=1\Leftrightarrow Q=-2\).
ĐKXĐ : \(y\ge\frac{4}{\sqrt{3}}\) hoặc \(y\le\frac{-4}{\sqrt{3}}\)
\(B=-\left|1-2x\right|-2\left|x-3\right|-\sqrt{3y^2-16}+2021\)
\(B=-\left(\left|1-2x\right|+\left|2x-6\right|\right)-\sqrt{3y^2-16}+2021\)
\(B\le-\left|1-2x+2x-6\right|-0+2021=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(1-2x\right)\left(2x-6\right)\ge0\left(1\right)\\3y^2-16=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}1-2x\ge0\\2x-6\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\ge3\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}1-2x\le0\\2x-6\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le3\end{cases}\Leftrightarrow}\frac{1}{2}\le x\le3}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(y^2=\frac{16}{3}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=\sqrt{\frac{16}{3}}\\y=-\sqrt{\frac{16}{3}}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{4}{\sqrt{3}}\\y=\frac{-4}{\sqrt{3}}\end{cases}}}\) ( nhận )
Vậy GTNN của \(B\) là \(2016\) khi \(\frac{1}{2}\le x\le3\) và \(y=\frac{4}{\sqrt{3}}\) hoặc \(y=\frac{-4}{\sqrt{3}}\)
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