Biết \(a^2+b^2=1;a^4+b^4=\frac{1}{2}.\). Tính \(a^{2020}+b^{2020}\)
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H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
MN
0
NK
1
AH
Akai Haruma
Giáo viên
21 tháng 8 2023
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$
$\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=4$
$\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$
$\Leftrightarrow 2+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$
$\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1$
$\Leftrightarrow \frac{a+b+c}{abc}=1$
$\Leftrightarrow a+b+c=abc$ (đpcm)
11 tháng 9 2023
Ta có: a+b+c=0\(\Leftrightarrow\)b+c=-a
Bình phương hai vế có: (b+c)2=a2
⇔ b2+2bc+c2=a2\(\Leftrightarrow\) b2+c2-a2=-2bc
Tương tự, ta có: c2+a2-b2=-2ca
a2+b2-c2=-2ab
→ A=\(-\dfrac{1}{2bc}-\dfrac{1}{2ca}-\dfrac{1}{2ab}=\dfrac{-\left(a+b+c\right)}{2abc}=0\)(vì a+b+c=0)
Vậy A=0
L
0
CT
0
Ta có:
\(a^2+b^2=1\Leftrightarrow\left(a^2+b^2\right)^2=1\Leftrightarrow a^4+b^4+2a^2b^2=1\)
\(\Leftrightarrow a^2b^2=\frac{1}{4}\Leftrightarrow b^2=\frac{1}{4a^2}\)
=> \(a^2+\frac{1}{4a^2}=1\Leftrightarrow4a^4-4a^2+1=0\Leftrightarrow\left(2a^2-1\right)^2=0\Leftrightarrow a^2=\frac{1}{2}\)
=> \(b^2=\frac{1}{2}\)
=> \(a^{2020}+b^{2020}=\left(a^2\right)^{1010}+\left(b^2\right)^{1010}=\left(\frac{1}{2}\right)^{1010}+\left(\frac{1}{2}\right)^{1010}=2.\frac{1}{2^{1010}}=\frac{1}{2^{2009}}\)