\(\dfrac{4}{9}-25x^2=0\)

\(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)

\(\left(\dfrac{2}{3}+5x\right)\left(\dfrac{2}{3}-5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}+5x=0\\\dfrac{2}{3}-5x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{2}{3}\\-5x=-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=\dfrac{2}{15}\end{matrix}\right.\)

\(\dfrac{4}{9-25x^2}=0\\ \Leftrightarrow\dfrac{4}{\left(3-5x\right)\left(3+5x\right)}=0\\ \Leftrightarrow\left(3-5x\right)\left(3+5x\right)=0\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\3+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Vậy....

x² - 25x⁴=0

<=>x²(1-25x²)=0

<=>x²[1²-(5x)²]=0

<=>x²(1-5x)(1+5x)=0

=>x²=0                =>x=0 

hoặc 1-5x=0         =>1=5x <=>x=0,2

hoặc 1+5x=0        =>1=-5x <=> x=-0,2

(x+2)+(4x+4)+(7x+6)+...+(25x+18)+(28x+20)=1560

(x+4x+7x+...+25x+28x)+(2+4+6+..+18+20)=1560

145x+110=1560

145x=1450

x=10

vậy........

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

a) \(\left(2x-3\right)\left(2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)