a) \(35x^9y^n=5.\left(7x^9y^n\right)\)

Để \(35x^9y^n⋮\left(-7x^7y^2\right)\)

\(\Rightarrow n\in\left\{0;1;2\right\}\)

b) \(5x^3-7x^2+x=3x\left(\dfrac{5}{3}x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)\)

Để \(\left(5x^3-7x^2+x\right)⋮3x^n\)

\(\Rightarrow3x\left(\dfrac{5}{3}x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)⋮3x^n\)

\(\Rightarrow n\in\left\{0;1\right\}\)

1) n=5, m=11

2) a) a>7

b)a<7

3) mình hơi rối mong bạn khá giải đáp bạn rõ hơn !!!

a) \(\left(5x^3-7x^2+x\right):3x^n\)

Để phép tính này chia hết thì

\(\left\{{}\begin{matrix}5x^3⋮3x^n\\-7x^2⋮3x^n\\x⋮3x^n\end{matrix}\right.\Rightarrow n\le1}\)

b) \(\left(13x^4y^3-5x^3y^3+6x^2y^2\right):5x^ny^n\)

Để phép tính này chia hết thì

\(\left\{{}\begin{matrix}13x^4y^3⋮5x^ny^n\\-5x^3y^3⋮5x^ny^n\\6x^2y^2⋮5x^ny^n\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}n\le4\\n\le3\end{matrix}\right.\\\left\{{}\begin{matrix}n\le3\\n\le3\end{matrix}\right.\\\left\{{}\begin{matrix}n\le2\\n\le2\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n\le3\\n\le3\\n\le2\end{matrix}\right.\Rightarrow n\le2}\)

Câu 1 :

\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:

\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)

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

Ta có:

\(\left(x+2\right)^2\ge0\forall x\)

Vậy pt vô nghiệm

Vậy:ko......

Câu 3:

\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)

\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)

Câu 4:

\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)

\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)

câu 6,

Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)

\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)

\(\Rightarrow-4x+3=8\)

\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)

Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

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

\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)

\(\Rightarrow7x^2=-x\)

\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).

Câu 2: \(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)\)

\(=\left(6x-9\right)\left(3x+2\right)-\left(2x+8\right)\left(4x-3\right)+36x-9x^2\)

\(=18x^2-27x+12x-18-8x^2-32x+6x+24+36x-9x^2\)

\(=x^2-5x+6\)

\(=x^2-2x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}+ 6\)

\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\)

Khi đó: \(\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}=0\)

\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{1}{2}\\x-\dfrac{5}{2}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\).

a) \(M\left(x\right)=-2x^5+5x^2+7x^4-5x+8+2x^5-7x^4-4x^2+6\)

\(=\left(-2x^5+2x^5\right)+\left(7x^4-7x^4\right)+\left(5x^2-4x^2\right)-9x+\left(8+6\right)\)

\(=x^2-9x+14\)

\(N\left(x\right)=7x^7+x^6-5x^3+2x^2-7x^7+5x^3+3\)

\(=\left(7x^7-7x^7\right)+x^6-\left(5x^3-5x^3\right)+2x^2+3\)

\(=x^6+2x^2+3\)

b) Đa thức M(x) có hệ số cao nhất là 1 

                                hệ số tự do là 14

                                bậc 2

 Đa thức N(x) có hệ số cao nhất là 1 

                            hệ số tự do là 3 

                            bậc 6

(2x + 3)² - (5x - 4)(5x - 4) = ( x + 5)² - (3x - 1)(7x + 2) - (x² - 1 +1)

<=> 4x² + 12x + 9 - ( 25x² - 16)= x² + 10x + 25 - (21x² + 6x - 7x - 2) -x²

<=> 4x² - 25x² - x² + 21x² + x² + 12x - 10x + 6x - 7x + 9 + 16 - 25 - 2 = 0

<=> x - 2 = 0

<=> x = 2

Vậy x = 2