(5x + 1)2 = 36/25
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1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)
2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)
3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)
4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)
\(F\left(x\right)=x^2-3x^3-\sqrt{25}+\frac{1}{2}x-\left(-3x^3+\frac{5x}{2}-\sqrt{36}\right)\)
=> \(F\left(x\right)=x^2-3x^3-5+\frac{1}{2}x+3x^3-\frac{5x}{2}+6\)
=> \(F\left(x\right)=x^2+\left(3x^3-3x^3\right)+\left(6-5\right)+\left(\frac{x}{2}-\frac{5x}{2}\right)\)
=> \(F\left(x\right)=x^2+1-2x\)
5x+35=25 -2 |x| +4=-20 4.13.(-125).(-25).8 -62.25+(-25).36
5x=25-35=-10 -2|x|=20-4=16 =(-25.4).(-125.8).13 =-25.(62+36)
x=-10/5=-2 |x|=16/-2=-8 =-100.-1000.13 =-25.98
Vậy x ko có giá trị nào do |x|=-8 =1300000 =-2450
1, \(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
vậy \(x\) \(\in\) {-3; 3}
5, 4\(x^2\) - 36 = 0
4.(\(x^2\) - 9) = 0
\(x^2\) - 9 = 0
(\(x\) - 3)(\(x\) + 3) = 0
\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; 3}
\(1,\\ b,=\left(x-6\right)\left(x+6\right)\\ 3,\\ x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
\(a,5x-5^2=3x+25\)
\(5x-3x=25+5^2\)
\(2x=50\)
\(x=50:2\)
\(x=25\)
\(b,3x+69:23=2x-108:36\)
\(3x+3=2x-3\)
\(3x-2x=-3-3\)
\(x=-6\)
Học tốt
\(\left(5x+1\right)^2=\dfrac{36}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{5}\\5x+1=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{1}{5}\\5x=-\dfrac{11}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{25}\\x=-\dfrac{11}{25}\end{matrix}\right.\)