\(a,25x^2-1=15\)\(< =>x^2=\dfrac{16}{25}< =>x=\pm\dfrac{4}{5}\)

\(b,\left(x-4\right)^2-\left(5x+2\right)^2=0\)\(< =>\left(-4x-6\right)\left(6x-2\right)=0\)

\(< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(c,\left(x-1\right)\left(x-9\right)=0< =>\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)

a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)

\(\Leftrightarrow x^3-27-x^3+3x=0\)

\(\Leftrightarrow x=9\)

b: Ta có: \(8x^4+x=0\)

\(\Leftrightarrow x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

+Ta có a= -4; b= -3 ; c= 9 và a²+ b²- c= 16+ 9 - 9 = 16> 0

Suy ra (C) là đường tròn tâm I( -4; -3) và R= 4

Vậy B; C đúng.

+Thay O vào (C) ta có: 0²+ 0²+ 8.0+ 6.0 + 9= 0 vô lí . Vậy A đúng.

+Thay M( -1; 0) vào (C) ta có: (-1) ²+ 0²+ 8.(-1) + 6.0 + 9= 0 ( vô lý). Vậy D sai.

Chọn D.

\(a.x^2-7x-3x+21=0\Leftrightarrow\left(x^2-7x\right)-\left(3x-21\right)=0\)

\(\Leftrightarrow x\left(x-7\right)-3\left(x-7\right)=0\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

\(b.x^2+6x+2x+12=0\Leftrightarrow\left(x^2+6x\right)+\left(2x+12\right)=0\)

\(\Leftrightarrow x\left(x+6\right)+2\left(x+6\right)=0\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)

\(c.x^2+4x+5x+20=0\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\Leftrightarrow\left(x+5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)