a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

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

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

a. (3x - 1)² - (x + 3)² = 0

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

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

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

*Cách khác:

a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)

a) \(\left(x^3+3x^2-8x-20\right)\div\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(10x+20\right)\right]\div\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+x-10\right)\div\left(x+2\right)\)

\(=x^2+x-10\) \(\left(x\ne-2\right)\)

b,c bn tự đặt chia

không giúp

\(a,=2x^3y+2x^2y^2-6xy^3\\ b,=3x^3+6x^2-4x-8\\ c,=\left(4x^2+16x-20x-80+76\right):\left(x+4\right)\\ =\left[\left(x+4\right)\left(4x-20\right)+76\right]:\left(x+4\right)\\ =4x-20\left(dư.76\right)\\ d,=\left(x^4-x^2-x^3+x-2x^2+2\right):\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2-x-2\right):\left(x^2-1\right)\\ =x^2-x-2\)

em mới lp 7 =)))

Câu 1: C

Câu 2: B

a) 2x.(3x² – 5x + 3)        

=2x³-10x²+6x                                                                       

b(-2x-1).( x² + 5x – 3 ) – (x-1)³

=-2x³ - 10x² + 6x - x² - 5x + 3 - x³ + 3x² - 3x + 1

= -3x³ - 8x² - 2x + 4

   d) (6x⁵y² – 9x⁴y³ + 15x³y⁴) : 3x³y² 

=2x²-3xy+5y²

\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)

\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)

\(A=7x^3-10x^2+5x-1\)

\(B=-8x^4+7x^2+3\)

`M(x)-N(x)=`\((-x^3+3x^2+5x+4)-(x^3+4x^2+6x-5)\)

`=-x^3+3x^2+5x+4-x^3-4x^2-6x+5`

`= (-x^3-x^3)+(3x^2-4x^2)+(5x-6x)+(4+5)`

`= -2x^3-x^2-x+9`

`@`\(\text{dn inactive.}\)

Mình tưởng bạn thông minh nên bt làm rồi.