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\(8^x:2^x=16^{2017}\)
\(\Leftrightarrow\left(8:2\right)^x=16^{2017}\)
\(\Leftrightarrow4^x=16^{2017}\)
\(\Leftrightarrow4^x=\left(4^2\right)^{2017}\)
\(\Leftrightarrow4^x=4^{4034}\)
\(\Leftrightarrow x=4034\)
Vậy ....
sua lai bai cua minh
Neu \(\left(x-2017\right)^2=1\\ =>x-2017=1\\ =>x=2018\)
Vay \(25=8\left(x-2017\right)^2+y^2\\ =>25=8+y^2\\ =>y^2=17\left(loai\right)\)(do x;y \(\in N\))
Vay \(x=2017;y=5\)
Ta co
\(25-y^2=8\left(x-2017\right)^2\\ =>25=8\left(x-2017\right)^2+y^2\)
Do
\(8\left(x-2017\right)^2\le25\\ =>\left(x-2017\right)^2\le\frac{25}{8}\)
\(=>\left(x-2017\right)^2\in\left\{0;1\right\}\)
Neu
\(\left(x-2017\right)^2=0\\ x-2017=0\\ x=2017\)
Vay:
\(25=8\left(x-2017\right)^2+y^2\\ =>25=y^2\\ =>y=5\)
Neu
\(\left(x-2017\right)^2=1\\ =>x-2017=1\\ =>x=2018\)
Vay:
\(25=8\left(x-2017\right)^2+y^2\\ =>25=1+y^2\\ =>y^2=24\)(loai do x;y \(\in N\))
Vay x=2017 ; y=5
1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)
\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)
\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)
\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)
\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)
3) Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)
\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)
\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)
\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)
\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)
\(B< A\)
( x - 2 ) ^2 = 1^2 hoặc ( x - 2 )^2 = -1^2
=> x - 2 = 1 hoặc x - 2 = -1
Ta có : x - 2 = 1 => x = 2 + 1 => x = 3
x - 2 = - 1 => x = - 1 + 2 => x = 1
( 2x -1 )^3 = -8
=> (2x-1)^3 = -2^3
=> 2x-1 = -2 => 2x = -2+1 => 2x = -1 => x = -1 :2 => x = -1/2
(x+1/2)^2 = 1/16
=> (x+1/2)^2 = 1/8^2 hoặc (x+1/2)^2 = -1/8^2
=> x+1/2 = 1/8 hoặc x+1/2 = -1/8
Ta có : x+1/2 = 1/8
x= 1/8 - 1/2
x = 2/16 - 8/16
x = -6/16 = -3/8
x + 1/2 = -1/8
x = -1/8 - 1/2
x = -2/16 -8/16
x= -10/16 = -5/8
* ^ là mũ nhé bạn :))
Ta có : \(8^x:2^x=16^{2017}\Leftrightarrow2^{3x}:2^x=2^{4.2017}\Leftrightarrow2^{2x}=2^{8086}\Rightarrow2x=8086\Rightarrow x=4043\)
Vậy x =4043