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

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

a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)

b) \(x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

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

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

a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(x^3-25x=0\)

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

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

a. (3x + 5)(7 - 2x) + 6x(x + 4) = 0

<=> 21x - 6x² + 35 - 10x + 6x² + 24x = 0

<=> -6x² + 6x² + 21x - 10x + 24x = -35

<=> 35x = -35

<=> x = \(\dfrac{-35}{35}=-1\)

b. x³ - 25x = 0

<=> x(x² - 5²)

<=> x(x + 5)(x - 5) = 0

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

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

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

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

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

Ta có:

\(0.25x^3+x^2+x=0\)

\(\Leftrightarrow x^3+4x^2+4x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

\(0,25x^3+x^2+x=0\)

\(x\left(0,25x^2+x+1\right)=0\)

\(x\left[\left(0,5x\right)^2+2\cdot0,5x\cdot1+1^2\right]=0\)

\(x\left(0,5x+1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\0,5x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Vậy.....

Lời giải:
$x^3-x^2-25x+25=0$

$\Leftrightarrow x^2(x-1)-25(x-1)=0$

$\Leftrightarrow (x-1)(x^2-25)=0$

$\Leftrightarrow (x-1)(x-5)(x+5)=0$

$\Leftrightarrow x-1=0$ hoặc $x-5=0$ hoặc $x+5=0$

$\Leftrightarrow x=1$ hoặc $x=\pm 5$

Đáp án D

x³-25x=0

=> x(x²-25)=0

=> x(x²-5²)=0

=> x(x-5)(x+5)=0

=> x=0            hoặc x-5=0         hoặc x+5=0

=> x=0            hoặc x=5             hoặc x=-5

x³-25x=0

<=>x.(x²-25)=0

<=>x.(x-5)(x+5)=0

<=>x=0 hoặc x-5=0 hoặc x+5=0

<=>x=0 hoặc x=5 hoặc x=-5

x³-25x=0 ( x>=0) => x*(x² - 25) =0

Xét 2 trường hợp => x = 0;5

\(1,A⋮B\Leftrightarrow x^3-3x^2-ax+3=\left(x-1\right)\cdot a\left(x\right)\)

Thay \(x=1\)

\(\Leftrightarrow1-3-a+3=0\\ \Leftrightarrow a=1\)

\(2,A⋮B\Leftrightarrow3x^3-16x^2+25x+a=\left(x^2-4x+3\right)\cdot b\left(x\right)\\ \Leftrightarrow3x^3-16x^2+25x+a=\left(x-3\right)\left(x-1\right)\cdot b\left(x\right)\)

Thay \(x=1\)

\(\Leftrightarrow3-16+25+a=0\\ \Leftrightarrow a=-12\)

Thay \(x=3\)

\(\Leftrightarrow3\cdot27-16\cdot9+25\cdot3+a=0\\ \Leftrightarrow81-144+75+a=0\\ \Leftrightarrow12+a=0\Leftrightarrow a=-12\)

Vậy \(a=-12\)

Đáp án : D

b: Ta có: \(6^{2x-1}=216\)

\(\Leftrightarrow2x-1=3\)

hay x=2

c: Ta có: \(x^3=25x\)

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

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