Cho -2020a > -2020b. Khi đó?
A. a < b
B. a > b
C. a = b
D. Cả A, B, C đều sai.
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Áp dụng t/c dttsbn:
\(\dfrac{a+b+c-2020d}{d}=\dfrac{b+c+d-2020a}{a}=\dfrac{c+d+a-2020b}{b}=\dfrac{d+a+b-2020c}{c}=\dfrac{3\left(a+b+c+d\right)-2020\left(a+b+c+d\right)}{a+b+c+d}=-2017\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c-2020d=-2017d\\b+c+d-2020a=-2017a\\c+d+a-2020b=-2017b\\d+a+b-2020c=-2017c\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b+c=3d\\b+c+d=3a\\c+d+a=3b\\d+a+b=3c\end{matrix}\right.\Rightarrow a=b=c=d\)
\(F=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\\ F=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=4\)
a) Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{c}=\frac{b}{d}=\frac{2020b}{2020d}=\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)
=> \(\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)
=> \(\frac{a+2020b}{a-2020b}=\frac{c+2020d}{c-2020d}\)
b) \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Áp dụng dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
=> \(\frac{a}{b}=\frac{a+c}{b+d}\Rightarrow\frac{a}{a+c}=\frac{b}{b+d}\)
=> \(\frac{2020a}{2020\left(a+c\right)}=\frac{b}{b+d}\)
=> \(\frac{2020\left(a+c\right)}{2020a}=\frac{b+d}{b}\)
c) \(2a+3c\left(b+d\right)=\left(a+c\right)\left(2b+3d\right)\)
Câu c sai đề.
A = 2020 (a -b )
thay a-b =10 vào biểu tức A lúc đó ta đuợc :
A = 2020 x 10
=20200
A = 2020a - 2020b
= 2020 . ( a - b )
ta có a - b = 10 => 2020 . 10
= 20200
\(\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(c+a\right)}\ge\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}=\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\frac{a}{a+\sqrt{2020a+bc}}\le\frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tương tự: \(\frac{b}{b+\sqrt{2020b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\) ; \(\frac{c}{c+\sqrt{2020c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Cộng vế với vế: \(P\le1\)
Dấu "=" xảy ra khi \(a=b=c=...\)
\(\sqrt{2020a+\frac{\left(b-c\right)^2}{2}}\le\sqrt{2020a+\frac{\left(b+c\right)^2}{2}}=\sqrt{2020a+\frac{\left(1010-a\right)^2}{2}}\)
\(=\sqrt{\frac{1}{2}\left(a^2+2020a+1010^2\right)}=\frac{1}{\sqrt{2}}\left(a+1010\right)\)
=> \(VT\le\frac{1}{\sqrt{2}}\left(a+b+c+3.1010\right)=2020\sqrt{2}\)
Dấu "=" xảy ra khi a=1010;b=0;c=0 và các hoán vị
Ta có:
\(\frac{a}{3}=\frac{b}{5}=\frac{c}{7}\Rightarrow\left\{{}\begin{matrix}a=3k\\b=5k\\c=7k\end{matrix}\right.\)
\(\Rightarrow\frac{2019b-2020a}{2019c-2020b}=\frac{2019.5k-2020.3k}{2019.7k-2020.5k}=\frac{4035k}{4033k}=\frac{4035}{4033}>\frac{4033}{4033}=1\)
Vậy \(\frac{2019b-2020a}{2019c-2020b}>1\left(đpcm\right)\)
Ta có: -2020a > -2020b
Û -2020. − 1 2020 a < -2020. − 1 2020 b Û a < b.
Đáp án cần chọn là: A