tìm x biết
52.x.5x+2:25=1252
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\(25^{2x}:5^x=125^2\)
\(\Rightarrow5^{4x}:5^x=\left(5^3\right)^2\)
\(\Rightarrow5^{4x-x}=5^6\)
\(\Rightarrow5^{3x}=5^6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
252\(x\) : 5\(x\) =1252
5\(^{4x}\) : 5\(^x\) = 56
5\(3x\) = 56
3\(x\) = 6
\(x\) = 2
dẫu . là dấu nhân nha mn.
mong mn giải giúp mình mình đang cần gấp
ĐKXĐ : \(x\ne\pm5\)
\(C=\dfrac{\left(x+2\right)\left(x-2\right)}{x^2-25}.\dfrac{x^2-25}{x^2+10}=\dfrac{x^2-4}{x^2+10}\)
\(C=2\Leftrightarrow x^2-4=2x^2+20\Leftrightarrow x^2=-24\left(vô-lí\right)\)
\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\left[\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\frac{x^2-\left(x^2-10x+25\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{10x-25}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{10x-25}+\frac{x}{5-x}\)
\(=\frac{1}{x-5}-\frac{x}{x-5}\)
\(=\frac{1-x}{x-5}=-\frac{x-1}{x-5}=-\frac{x-5+4}{x-5}=-1-\frac{4}{x-5}\)
Để P nguyên <=> x - 5 thuộc Ư(4) = {1;-1;2;-2;4;-4}
Ta có bảng:
x - 5 | 1 | -1 | 2 | -2 | 4 | -4 |
x | 6 | 4 | 7 | 3 | 9 | 1 |
Vậy....
1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)
\(\Leftrightarrow25x^2-1=25x^2-7x+15\)
\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)
2) \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)
\(\Leftrightarrow5x=0\Leftrightarrow x=0\)
\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)
Vậy \(x=12\).
(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12
Vậy x=12x=12.
cho tôi đúng đi