Tham khảo:

https://hoc24.vn/hoi-dap/question/278669.html

Ý 2 nè:

https://hoc24.vn/hoi-dap/question/97085.html

từ đó rồi tiếp tục nha bạn

M=x¹⁴ -(x+1)x¹³+(x+1)x¹²-(x+1)x¹¹+...+(x+1)x²-(x+1)x+x+1

M=x¹⁴-x¹⁴-x¹³+x¹³+x¹²-x¹²-x¹¹+...+x³+x²-x²+x+1

M=1

Vậy m=1 tai x=9

\(A=x^{10}-14x^9-x^9+14x^8+x^8-14x^7-x^7...-x+14+1\)

\(A=x^9\left(x-14\right)-x^8\left(x-14\right)+x^7\left(x-14\right)-...-x\left(x-14\right)+1\)

\(A=1\) (Do x=14)

7/5 x X + 3/8 x X= 9

X x (7/5 +3/8)=9

X x 71/40 = 9

X             = 9:71/40

X             = 360/71

     \(\frac{7}{15}X+\frac{3}{8}X=9\)

     \(\left(\frac{7}{15}+\frac{3}{8}\right)X=9\) 

\(\left(\frac{56}{120}+\frac{45}{120}\right)X=9\)

                  \(\frac{101}{120}X=9\)

                          \(X=9:\frac{101}{120}\)

                          \(X=9x\frac{120}{101}\)

                          \(X=\frac{1080}{101}\)

\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)

a: \(\Leftrightarrow3x\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ta có x=14 suy ra x+1=15

Do đó thay x+1 vào H(x), ta được:

H(14)= x¹⁰ - (x+1)x⁹ +(x+1)x⁸-(x+1)x⁷+...+ (x+1)x² - (x+1)x + x+1

H(14)= x^10 - x^10 -x^9 +x^8- x^8-x^7 +....+ x^3 +x^2 -x^2-x+x +1

Hay H(14)=1

Đặt Q(x) = x⁹ - x⁸ + x⁷ - ... + x - 1 thì (x + 1) * Q(x) = (x + 1) * (x⁹ - x⁸ + x⁷ - ... + x - 1) = x¹⁰ - 1 \(\Rightarrow\left(14+1\right)\cdot Q\left(14\right)=14^{10}-1\)

Dễ thấy: H(x) = x¹⁰ - 15* Q(x) \(\Rightarrow H\left(14\right)=14^{10}-\left(14^{10}-1\right)=1\)

Xin cậu !