Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{b^2k^{2022}+b^{2022}}{d^{2022}k^{2022}+d^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)
\(\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\dfrac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)
=>\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)
Lời giải:
$b^2=ac\Rightarrow \frac{b}{a}=\frac{c}{b}$
Đặt $\frac{b}{a}=\frac{c}{b}=k\Rightarrow b=ak; c=bk$
Khi đó:
$\frac{a^{2022}+b^{2022}}{b^{2022}+c^{2022}}=\frac{a^{2022}+(ak)^{2022}}{b^{2022}+(bk)^{2022}}$
$=\frac{a^{2022}(1+k^{2022})}{b^{2022}(1+k^{2022})}=\frac{a^{2022}}{b^{2022}} (1)$
Và:
$(\frac{a+b}{b+c})^{2022}=(\frac{a+ak}{b+bk})^{2022}$
$=[\frac{a(k+1)}{b(1+k)}]^{2022}=(\frac{a}{b})^{2022}=\frac{a^{2022}}{b^{2022}}(2)$
Từ $(1); (2)$ ta có đpcm.
TA có: \(3a-b=\frac12\left(a+b\right)\)
=>2(3a-b)=a+b
=>6a-2b=a+b
=>5a=3b
=>a=0,6b
\(C=\frac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\frac{\left(0,6b\right)^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\frac{\left(0,6\right)^{2022}\left(b^{2022}+5^{2022}\right)}{b^{2022}+5^{2022}}=\left(0,6\right)^{2022}=\left(\frac35\right)^{2022}\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
sửa đề: Cho \(\frac{a}{b}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\left(\frac{a-b}{c-d}\right)^{2022}=\left(\frac{bk-b}{dk-d}\right)^{2022}=\left\lbrack\frac{b\left(k-1\right)}{d\left(k-1\right)}\right\rbrack^{2022}=\left(\frac{b}{d}\right)^{2022}\)
\(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}\)
\(=\frac{\left(bk\right)^{2022}+b^{2022}}{\left(dk\right)^{2022}+d^{2022}}=\frac{b^{2022}\left(k^{2022}+1\right)}{d^{2022}\left(k^{2022}+1\right)}=\frac{b^{2022}}{d^{2022}}=\left(\frac{b}{d}\right)^{2022}\)
Do đó; \(\left(\frac{a-b}{c-d}\right)^{2022}=\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(bk\right)^{2022}+b^{2022}}{\left(dk\right)^{2022}+d^{2022}}=\frac{b^{2022}\left(k^{2022}+1\right)}{d^{2022}\left(k^{2022}+1\right)}=\frac{b^{2022}}{d^{2022}}\)
\(\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\frac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\frac{\left\lbrack b\left(k+1\right)\right\rbrack^{2022}}{\left\lbrack d\left(k+1\right)\right\rbrack^{2022}}=\frac{b^{2022}}{d^{2022}}\)
Do đó: \(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)