Tìm y, biết:
a) y^2015=y^2020
b) ( 2y-1)^50 - ( 2y-1)^1
c) (\(\frac{y}{3}\)- 5)\(^{2020}\)= (\(\frac{Y}{3}\)- 5)\(^{2008}\)
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\(a,\Leftrightarrow y^{200}-y=y\left(y^{199}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y^{199}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\)
Vậy ..
\(b,\Leftrightarrow y^{2010}-y^{2008}=y^{2008}\left(y^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^{2008}=0\\y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=-1\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)=\left(2y-1\right)\left(\left(2y-1\right)^{49}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2y-1=0\\\left(2y-1\right)^{49}=1\end{matrix}\right.\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy ..
\(d,\Leftrightarrow\left(\dfrac{y}{3}-5\right)^{2008}\left(\left(\dfrac{y}{3}-5\right)^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{y}{3}-5\right)^{2008}=0\\\left(\dfrac{y}{3}-5\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{y}{3}-5=0\\\dfrac{y}{3}-5=1\\\dfrac{y}{3}-5=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=15\\y=18\\y=12\end{matrix}\right.\)
Vậy ..
\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)
Vậy \(S=\varnothing\)
b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) y^200 = y
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}\)
b) y^2008 = y^2010
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}\)
c) (2y - 1)^50 = 2y - 1
\(\Leftrightarrow\orbr{\begin{cases}2y-1=1\\2y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=1\\y=\frac{1}{2}\end{cases}}\)
d) (y/3 - 5)^2000= y/3 -5
\(\Leftrightarrow\orbr{\begin{cases}\frac{y}{3}-5=1\\\frac{y}{3}-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=18\\y=15\end{cases}}\)
( x - 1 )5 = 32
Mà 25 = 32
=> x - 1 = 2
=> x = 2 + 1
=> x = 3
Vậy x = 3
( x - 1 )5 = 32 y200 = y
( x - 1 )5 = 25 => y = 1
=> x - 1 = 2
x = 2 + 1
x = 3
Vậy x = 3
a) \(y^{2015}=y^{2020}\)
\(\Leftrightarrow y^{2020}-y^{2015}=0\)
\(\Leftrightarrow y^{2015}.\left(y^5-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y^{2015}=0\\y^5-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)
Vậy ...
b) \(\left(2y-1\right)^{50}=\left(2y-1\right)^1\)
\(\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)^1=0\)
\(\Leftrightarrow\left(2y-1\right)^1.\left[\left(2y-1\right)^{49}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2y-1\right)^1=0\\\left(2y-1\right)^{49}-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{2}\\y=1\end{cases}}\)
Vậy...