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\(2.3^x-3^{x-1}=405\)
\(2.3^x-3^x.3^{-1}=405\)
3^x(2+3)=405
3^x=81
x=4
dung cho minh k nha
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
bạn thử viết lại đề bài cho rõ hơn được không, mình sẽ giải giúp bạn.
a, đầu bài sai hay sao đó bn
b,\(\dfrac{2}{3}\).3x+1 - 7.3x = -405
3x(\(\dfrac{2}{3}\).3 - 7) = -405
3x . (-5) = -405
3x = 81 mà 81 = 34 suy ra x = 4
\(\dfrac{2}{3}.3^{x+1}-7.3^x=-405\)
\(\Rightarrow\dfrac{2}{3}.3^x.3^1-7.3^x=-405\)
\(\Rightarrow2.3^x-7.3^x=-405\)
\(\Rightarrow3^x.\left(2-7\right)=-405\)
\(\Rightarrow3^x.-5=-405\)
\(\Rightarrow3^x=\dfrac{-405}{-5}\)
\(\Rightarrow3^x=81\)
Vì \(3^4=81\) nên \(x=4\)
\(\dfrac{2}{3}.3^{x+1}-7.3^x=-405\)
\(\Leftrightarrow\dfrac{2}{3}.3.3^x-7.3^x=-405\)
\(\Leftrightarrow3^x\left(2-7\right)=-405\)
\(\Leftrightarrow3^x.\left(-5\right)=-405\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Leftrightarrow x=4\)
Vậy..
a) \(\frac{2}{3}.3^{x+1}-7.3^x=-405\)
=> \(\frac{2}{3}.3^x.3-7.3^x=-405\)
=> \(2.3^x-7.3^x=-405\)
=> \(3^x\left(2-7\right)=-405\)
=> \(3^x.\left(-5\right)=-405\)
=> \(3^x=-405:\left(-5\right)\)
=> \(3^x=81\)
=> \(x=4\)
b) \(\frac{3}{1-2x}=\frac{-5}{3x-2}\)
=> \(3\left(3x-2\right)=-5\left(1-2x\right)\)
=> \(9x-6=-5+10x\)
=> \(9x-10x=-5+6\)
=> \(-x=1\)
=> \(x=-1\)
Ta có: \(2.3^x-405=3^{x-1}\)
\(\Leftrightarrow2.3^x-3^x:3^1=405\)
\(\Leftrightarrow3^x.\left(2-\frac{1}{3}\right)=405\)
\(\Leftrightarrow3^x.\frac{5}{3}=405\)
\(\Leftrightarrow3^x=81.3\)
\(\Leftrightarrow3^x=3^4.3\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x\in\left\{5\right\}\)