Cho b^2=ác. CM:a^2022+b^2022/b^2022+c^2022=(a+b/b+c)^2022
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TN
1
30 tháng 12 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{b^2k^{2022}+b^{2022}}{d^{2022}k^{2022}+d^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)
\(\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\dfrac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\left(\dfrac{b}{d}\right)^{2022}\)
=>\(\dfrac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\dfrac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)
31 tháng 7 2023
a^2+b^2+c^2=ab+bc+ac
=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0
=>a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0
=>(a-b)^2+(b-c)^2+(a-c)^2=0
=>a=b=c
\(T=\dfrac{a^{2022}+a^{2022}+a^{2022}}{\left(3a\right)^{2022}}=\dfrac{3}{3^{2022}}=\dfrac{1}{3^{2021}}\)
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
9 tháng 5 2022
\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
30 tháng 6 2023
3a-b=1/2(a+b)
=>6a-2b=a+b
=>5a=3b
=>a/3=b/5=k
=>a=3k; b=5k
\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)
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.