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\(\dfrac{4}{9}-25x^2=0\)
\(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\left(\dfrac{2}{3}+5x\right)\left(\dfrac{2}{3}-5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}+5x=0\\\dfrac{2}{3}-5x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{2}{3}\\-5x=-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=\dfrac{2}{15}\end{matrix}\right.\)
\(\dfrac{4}{9-25x^2}=0\\ \Leftrightarrow\dfrac{4}{\left(3-5x\right)\left(3+5x\right)}=0\\ \Leftrightarrow\left(3-5x\right)\left(3+5x\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\3+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy....
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
\(x^2\left(x+1\right)+2x\left(x+1\right)=0\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\left\{{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
a, \(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)
=> ( x + 1 ) ( \(x^2\) + 2x ) = 0
=> ( x + 1 ) x (x + 2 ) = 0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\)
=> \(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
=> \(\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=\dfrac{-2}{15}\end{matrix}\right.\)
A) x3-6x2+12x-8=0
<=>(x-2)3=0
<=>x-2=0
<=>x=2
B)4(x-3)2 -(2x-1)(2x+1)=13
<=>4(x2-6x+9)-4x2+1=13
<=>4x2-24x+36-4x2+1=13
<=>-24x+37=13
<=>24x=37-13
<=>24x=24
<=>x=1
C)25x2-6(x+1)2=0
<=>(5x-\(\sqrt{6}\left(x+1\right)\))(5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}\left(x+1\right)\)=0 hoặc 5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}x-\sqrt{6}\)=0 <=>5x+\(\sqrt{6}x+\sqrt{6}\)=0
<=>x(5-\(\sqrt{6}\))=\(\sqrt{6}\) <=>x(5+\(\sqrt{6}\))=\(-\sqrt{6}\)
<=>x=\(\frac{\sqrt{6}}{5-\sqrt{6}}\) <=>x=\(\frac{-\sqrt{6}}{5+\sqrt{6}}\)
Rút gọn C=(4+2A+A^2).(4-A^2).(4-2a+a^2) GIẢI GIÚP MIK ĐI
\(\dfrac{4}{9-25x^2}\\ =\dfrac{4}{\left(3-5x\right)\left(3+5x\right)}\)