Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(b^2=ac\)
=>\(\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\)
=>a=bk; b=ck
=>\(a=ck\cdot k=ck^2;b=ck\)
\(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{\left(ck^2+ck\right)^{2021}}{\left(ck+c\right)^{2021}}=\frac{\left\lbrack ck\left(k+1\right)\right\rbrack^{2021}}{\left\lbrack c\left(k+1\right)\right\rbrack^{2021}}=k^{2021}\)
\(\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}=\frac{\left(ck^2\right)^{2021}+\left(ck\right)^{2021}}{\left(ck\right)^{2021}+c^{2021}}\)
\(=\frac{c^{2021}\cdot k^{2021}\left(k^{2021}+1\right)}{c^{2021}\left(k^{2021}+1\right)}=k^{2021}\)
Do đó: \(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)
\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)
Ta có: \(b^2=ac\)
=>\(\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\)
=>a=bk; b=ck
=>\(a=ck\cdot k=ck^2;b=ck\)
\(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{\left(ck^2+ck\right)^{2021}}{\left(ck+c\right)^{2021}}=\frac{\left\lbrack ck\left(k+1\right)\right\rbrack^{2021}}{\left\lbrack c\left(k+1\right)\right\rbrack^{2021}}=k^{2021}\)
\(\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}=\frac{\left(ck^2\right)^{2021}+\left(ck\right)^{2021}}{\left(ck\right)^{2021}+c^{2021}}\)
\(=\frac{c^{2021}\cdot k^{2021}\left(k^{2021}+1\right)}{c^{2021}\left(k^{2021}+1\right)}=k^{2021}\)
Do đó: \(\frac{\left(a+b\right)^{2021}}{\left(b+c\right)^{2021}}=\frac{a^{2021}+b^{2021}}{b^{2021}+c^{2021}}\)
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\)
\(=3+Q\)
Suy ra \(3+Q=1\Leftrightarrow Q=-2\).
Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
=> \(\left(\frac{a}{c}\right)^{2021}=\left(\frac{b}{d}\right)^{2021}=\left(\frac{a-b}{c-d}\right)^{2021}\)
=> \(\frac{a^{2021}}{c^{2021}}=\frac{b^{2021}}{d^{2021}}=\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)
=>\(\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)(đpcm)