Tìm x biết
1253x = 514
25x
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a)
(x - 3) : 2 = 5 14 : 5 12
(x - 3) : 2 = 5 2
(x - 3) : 2 = 25
(x - 3) = 25.2
x = 50 + 3
x = 53
a) (x - 3) : 2 = 5 14 : 5 12
(x - 3) : 2 = 5 2
(x - 3) : 2 = 25
(x - 3) = 25.2
x = 50 + 3
x = 53
b) 4x + 3x = 30 – 20 : 10
7x = 30 - 2
7x = 28
x = 28 : 7
x = 4
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
\(\Leftrightarrow\left(5x-7\right)\left(5x+7-x-3\right)=0\)
\(\Leftrightarrow\left(5x-7\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)
(x+1)+(x+4)+...+(x+46)=514
(x+x+x+...+x)+(1+4+7+...+46)=514
16x+ [46+1]*[( 46-1):3+1]:2=514
16x+376=514
16x=514-376
16x=138
x=138:16
x=8,6
Vậy x=8,6
1) ĐKXĐ: \(x\ge-2\)
\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)
\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
2) ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)
\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)
\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)
5) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)
\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)
Vậy \(S=\varnothing\)
\(\frac{125^{3x}}{25^x}=5^{14}=>\frac{\left(5^3\right)^{3x}}{\left(5^2\right)^x}=5^{14}=>\frac{5^{9x}}{5^{2x}}=5^{14}=>5^{7x}=5^{14}=>7x=14=>x=2\)