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(x-3)6 = (x-3)4
<=> (x-3)4.(x-3)2 - (x-3)4 = 0
<=> (x-3)4.[(x-3)2 - 1] = 0
<=> \(\left[\begin{array}{nghiempt}\left(x-3\right)^4=0\Rightarrow x=3\\\left(x-3\right)^2-1=0\Rightarrow x\in\left\{2;4\right\}\end{array}\right.\)
Vậy...
\(\Leftrightarrow2\left(2x-3\right)=5\left(x+4\right)\)
=>5x+20=4x-6
=>x=-26
\(\Leftrightarrow5^x\cdot125-5^x=175\)
\(\Leftrightarrow5^x\cdot124=175\)
\(\Leftrightarrow5^x=\dfrac{175}{124}\)
\(\left(3x-2\right)^3=125\)
\(\Leftrightarrow\left(3x-2\right)^3=5^3\)
\(\Leftrightarrow\left(3x-2\right)=5\)
\(\Leftrightarrow3x=7\)
\(\Leftrightarrow x=\frac{7}{3}\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
sai đề