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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
Ta có 24 mũ 3 : 12 mũ 3 = (24:12) mũ 3 = 2 mũ
25 mũ 6 : 25 mũ 4 = 25 mũ (4-2) = 25 mũ 2
81 mũ 2 : 9 mũ 2 = (81:9) mũ 2 = 9 mũ 2
32 mũ 4 : 2 mũ 6 = 16834
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+4\right)\left(x+1\right)}=\dfrac{3}{10}\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=10\)
\(\Leftrightarrow x^2+5x-6=0\)
=>(x+6)(x-1)=0
=>x=-6 hoặc x=1
\(12^8:2^8< =6^x< =\left(6^3\right)^3\)
<=> \(1679616< =6^x< =6^9\)
<=> \(6^8< =6^x< =6^9\)
<=> \(8< =x< =9\)
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
\(12^2.2.12^3.6\)
\(=\left(12^2.12^3\right).\left(2.6\right)\)
\(=12^5.12\)
\(=12^6\)
2985984 nha b