\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=> \(3^{n-1}+5.3^{n-1}=162\)
<=> \(3^{n-1}\left(1+5\right)=162\)
<=> \(3^{n-1}.6=162\)
<=> \(3^{n-1}=162:6\)
<=> \(3^{n-1}=27\)
<=> \(3^{n-1}=3^3\)
<=> n - 1 = 3
<=> n = 3 + 1 = 4
Câu 1
a) Từ gt=>\(\hept{\begin{cases}x-5=1-3x\\x-5=3x-1\end{cases}}\)
<=>\(\hept{\begin{cases}4x=6\\2x=-4\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
b) Ta có: \(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0,\forall x\in R\\\left(2y+1\right)^{200}\ge0,\forall x\in R\end{cases}}\)
Kết hợp với đề bài => \(\hept{\begin{cases}3x-1=0\\2y+1=0\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)
Bài 2
\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=>\(3^{n-1}+5.3^{n-1}=162\)
<=>\(6.3^{n-1}=162\)
<=>\(3^{n-1}=27=3^3\)
<=>\(n-1=3\)
<=>\(n=4\)
b, Dấu = khi \(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)
\(pt< =>6.3^{n-1}=162< =>3^{n-1}=3^3< =>n=4\)
a) \(\left(x-5\right)^2=\left(1-3x\right)^2\)
<=> \(\left(x-5\right)^2-\left(1-3x\right)^2=0\)
<=> \(\left(x-5-1+3x\right)\left(x-5+1-3x\right)=0\)
<=> \(\left(4x-6\right)\left(-2x-4\right)=0\)
<=> \(-4\left(2x-3\right)\left(x+2\right)=0\)
<=> \(\left(2x-3\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}2x-3=0\\x+2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
Bài 1:
a) \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(1-3x\right)^2=0\)
\(\Leftrightarrow\left(4x-6\right)\left(-2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-6=0\\-2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
b) Ta có: \(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0\\\left(2y+1\right)^{200}\ge0\end{cases}\left(\forall x,y\right)}\)
\(\Rightarrow\left(3x-1\right)^{100}+\left(2y+1\right)^{200}\ge0\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(3x-1\right)^{100}=0\\\left(2y+1\right)^{200}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)
Bài 2:
Ta có:\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
\(\Leftrightarrow3^{n-1}+5.3^{n-1}=162\)
\(\Leftrightarrow6.3^{n-1}=162\)
\(\Leftrightarrow3^{n-1}=27=3^3\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
Cảm ơn mọi người nhiều!!!!
a) ( x - 5 )2 = ( 1 - 3x )2
<=> ( x - 5 )2 - ( 1 - 3x )2 = 0
<=> [ ( x - 5 ) - ( 1 - 3x ) ][ ( x - 5 ) + ( 1 - 3x ) ] = 0
<=> [ 4x - 6 ][ -2x - 4 ] = 0
<=> \(\orbr{\begin{cases}4x-6=0\\-2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
b) \(\left(3x-1\right)^{100}+\left(2y+1\right)^{200}\le0\)
\(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0\forall x\\\left(2y+1\right)^{200}\ge\forall y\end{cases}\Rightarrow}\left(3x-1\right)^{100}+\left(2y+1\right)^{200}\ge0\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}3x-1=0\\2y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}}\)
\(\frac{1}{3}\cdot3^n+5\cdot3^{n-1}=162\)
\(\Leftrightarrow3^{n-1}+5\cdot3^{n-1}=162\)
\(\Leftrightarrow3^{n-1}\left(1+5\right)=162\)
\(\Leftrightarrow3^{n-1}=27\)
\(\Leftrightarrow3^{n-1}=3^3\)
\(\Leftrightarrow n-1=3\Rightarrow n=4\)
