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Vì \(\left(x-2\right)^4\ge0\forall x\)dấu "=" xảy ra \(\Leftrightarrow\)x-2=0 \(\Leftrightarrow\)x=2
\(\left(2y-1\right)^{2014}\ge0\forall y\)Dấu "=" xảy ra \(\Leftrightarrow\)2y - 1=0 \(\Leftrightarrow y=\frac{1}{2}\)
\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2014}\ge0\)
Kết hợp với điều kiện đề bài \(\left(x-1\right)^4+\left(2y-1\right)^{2014}\le0\), ta được:
\(\left(x-2\right)^4+\left(2y-1\right)^{2014}=0\)
Vậy x = 2; \(y=\frac{1}{2}\)
Thay x=2; \(y=\frac{1}{2}\)vào M, ta có:
\(M=21.2^2.\frac{1}{2}+4.2.\left(\frac{1}{2}\right)^2\)
\(=21.4.\frac{1}{2}+4.2.\frac{1}{4}\)
\(=42+2=44\)
Vậy M=44
(2^x-8)^3=(4^x+2^x+5)^3-(4^x+13)^3
(2^x-8)^3=[(4^x+2^x+5)-(4^x+13)]*[(4^x... + (4^x+13)^2]
(2^x-8)^3=(2^x-8)*[(4^x+2^x+5)^2+(4^x+... + (4^x+13)^2]
2^x=8=>x=3
hoặc (2^x-8)^2=(4^x+2^x+5)^2+(4^x+2^x+5)(4^x+... + (4^x+13)^2
(4^x+2^x+5)^2 - (2^x-8)^2+(4^x+2^x+5)(4^x+13) + (4^x+13)^2=0
[(4^x+2^x+5)-(2^x-8)]*[(4^x+2^x+5)+(2^... + (4^x+3)*[(4^x+2^x+5)+(4^x+13)]=0
(4^x+13)*(4^x+2*2^x-3) + (4^x+3)*(2*4^x+2^x+18)=0
(4^x+13)[(4^x+2*2^x-3) + (2*4^x+2^x+18)]=0
4^x+13=0 (VN)
hoặc 3*4^x + 3*2^x +15=0
đặt t=2^x ( t>0)
t^2 + t + 5=0 ptvn
3n + 3 + 3n + 1 + 2n + 3 + 2n + 2
= 3n.33 + 3n.3 + 2n.23 + 2n.22
= 3n.(27 + 3) + 2n.(8 + 4)
= 3n.30 + 2n.12
= 3n.5.6 + 2n.2.6
= 6.(3n.5 + 2n.2) \(⋮\) 6
\(\frac{2016^{26}+2016^{24}+...+2016^4+2016^2}{2016^{24}+2016^{22}+...+2016^2+1}\) \(=\frac{2016^2.\left(2016^{24}+2016^{22}+...+2016^2+1\right)}{2016^{24}+2016^{22}+...+2016^2+1}\)
\(=\frac{2016^2}{1}=2016^2\)
Có: \(5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}=5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}\)
\(=3^{18}\cdot2^{29}\cdot\left(5\cdot2-3^2\right)=3^{18}\cdot2^{29}\)
Lại có: \(5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6=5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}=5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(=2^{19}\cdot3^{18}\cdot\left(5\cdot3-7\right)=2^{19}\cdot3^{18}\cdot2^3=2^{22}\cdot3^{18}\)
Vậy \(\frac{3^{18}\cdot2^{29}}{2^{22}\cdot3^{18}}=2^7=128\)
-3^4.4^4/2^2.6^2
=(-3x4)^4/(2x6)^2
=(-12)^4/12^2
=(-12)^2
=144
Bài giải
Ta có : \(3^{2014}\cdot81^{19}=3^{2014}\cdot\left(3^4\right)^{19}=3^{2014}\cdot3^{76}=3^{2090}\)
\(6^{60}\cdot3^{1955}=\left(2\cdot3\right)^{60}\cdot3^{1955}=2^{60}\cdot3^{60}\cdot3^{1955}=2^{60}\cdot3^{2015}\)