a)\(x^4-6x^2+2x+28\)

\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(3x-3\right)+25\)

\(=\left(x-1\right)\left(x^3+x^2-5x-3\right)+25\)

=> số dư là 25

b) Cách làm tương tự câu a nhé

\(\dfrac{x^4+2x^3+10x-25}{x^2+5}\)

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

\(=x^2+2x-5\)

Ta có: 

\(2x^3-5x^2+6x-15\)

\(=\left(2x^3-5x^2\right)+\left(6x-15\right)\)

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

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

\(\Rightarrow\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)=x^2+3\)

= x + 3 nha

\(=\left[x^2\left(2x-5\right)+3\left(2x-5\right)\right]:\left(2x-5\right)\\ =x^2+3\)

\(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\\ =\left[x^2\left(2x-5\right)+3\left(2x-5\right)\right]:\left(2x-5\right)\\ =\left[\left(2x-5\right)\left(x^2+3\right)\right]:\left(2x+5\right)=x^2+3\)

3: \(\Leftrightarrow a-15=0\)

hay a=15

=0 bạn nha

a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

\(\Leftrightarrow x^2+x-8=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)