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\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\)
\(=3+Q\)
Suy ra \(3+Q=1\Leftrightarrow Q=-2\).
Ta có :
Đặt \(\frac{a}{2019}\)= \(\frac{b}{2020}\)= \(\frac{c}{2021}\)= k
=> a = 2019k; b = 2020k; c = 2021k
M = 4(a-b).(b-c) - (c-a)
M = 4(2019k- 2020k). (2020k-2021k) - (2021k - 2019k)
M = 4.(-1)k.(-1)k - 2k
M = 4k2 - 2k
(Hình như mình thấy đề bạn có gì sai sai)
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 : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
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)
=> \(\frac{a_1}{a_2}=\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\)
<=> \(\left(\frac{a_1}{a_2}\right)^{2020}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}...\frac{a_{2020}}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
<=> \(\frac{a_1}{a_{2021}}=\left(\frac{a_1+a_2+a_3+...+a_{2020}}{a_2+a_3+a_4+...+a_{2021}}\right)^{2020}\)
Áp dụng dãy tỉ số bằng nhau ta có:
: \(\frac{a}{2019}\)=\(\frac{b}{2020}\)=\(\frac{c}{2021}=\frac{c-a}{2021-1009}=\frac{a-b}{2019-2020}=\frac{b-c}{2020-2021}\)
=> \(\frac{c-b}{2}=\frac{a-b}{-1}=\frac{b-c}{-1}\)
=> \(\frac{\left(c-b\right)^2}{4}=\frac{\left(a-b\right)\left(b-c\right)}{1}\)
=> \(\left(c-a\right)^2=4\left(a-b\right)\left(b-c\right)\)
Đặt \(\frac{a}{2019}=\frac{b}{2020}=\frac{c}{2021}=y\)
\(\Rightarrow a=2019y;b=2020y;c=2021y\)
\(\Rightarrow\hept{\begin{cases}4\left(a-b\right)\left(b-c\right)=4\left(2019y-2020y\right)\left(2020y-2021y\right)=4\left(-y\right)\left(-y\right)=4y^2\\\left(c-a\right)^2=\left(2021y-2019y\right)^2=4y^2\end{cases}}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)( ĐPCM )
a,theo t/c dãy tỉ số = nhau ta có:
a/b=b/c=c/a=a+b+c/b+c+a=1
suy ra a=b=c mà a=2021
suy ra b=2021,c=2021
suy ra b+c=2021+2021=4042
b, a2021.b2021/c2021 mà a=b=c suy ra a2021.a2021/a2021=a2021 mà a=2021 suy ra a2021.b2021/c2021=20212021