Đặt \(P\left(n\right)=n^4-10n^3+35n^2-50n+24\)

Ta có \(P\left(n\right)=n^4-n^3-9n^3+9n^2+26n^2-26n-24n+24\)

\(P\left(n\right)=n^3\left(n-1\right)-9n^2\left(n-1\right)+26n\left(n-1\right)-24\left(n-1\right)\)

\(P\left(n\right)=\left(n-1\right)\left(n^3-9n^2+26n-24\right)\)

Đặt \(H\left(n\right)=n^3-9n^2+26n-24\). Khi đó \(P\left(n\right)=\left(x-1\right).H\left(n\right)\)

mà \(H\left(n\right)=n^3-9n^2+26n-24\)

\(H\left(n\right)=n^3-2n^2-7n^2+14n+12n-24\)

\(H\left(n\right)=n^2\left(n-2\right)-7n\left(n-2\right)+12\left(n-2\right)\)

\(H\left(n\right)=\left(n-2\right)\left(n^2-7n+12\right)\)

Dễ dàng thấy được \(n^2-7n+12=n^2-3n-4n+12=n\left(n-3\right)-4\left(n-3\right)=\left(n-3\right)\left(n-4\right)\)

Do đó \(H\left(n\right)=\left(n-2\right)\left(n-3\right)\left(n-4\right)\). Từ đó \(P\left(n\right)=\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\)

Mà đây chính là tích của 4 số liên tiếp, trong 4 số này luôn tồn tại một bội của 4, một bội của 3 và một số khác là bội của 2 nên \(P\left(n\right)⋮2.3.4=24\), và ta có đpcm

\(10+x^2+6x\)

\(=x^2+6x+10\)

\(=x^2+2x.3+3^2+1\)

\(=\left(x+3\right)^2+1>0\) ( đfcm )

10 + x² + 6x = x² + 2.x .3 + 3² + 1 = (x+3)² + 1 

(x+3)² ≥ 0 ∀ x ϵ R ⇔ (x+3)² + 1 ≥ 1 >0 ∀ x ϵ R

⇔ 10 + x² + 6x > 0 ∀ x ϵ R (đpcm)

`7x(x+1)=x+1`

`<=>7x(x+1)-(x+1)=0`

`<=>(x+1)(7x-1)=0`

`<=>` $\left[\begin{matrix} x=-1\\ x=\dfrac{1}{7}\end{matrix}\right.$

Vậy `S={-1;1/7}`

\(7x\left(x+1\right)=\left(x+1\right)\)

\(\Leftrightarrow7x\left(x+1\right)-\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\7x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-1;\dfrac{1}{7}\right\}\)

5 là số nguyên tố. Theo định lý Fermat nhỏ

\(5^{2017}-5\equiv0\) (mod 2017)

\(\Rightarrow5^{2021}=5^{2017}.5^4=\left(5^{2017}-5+5\right).5^4=\)

\(=5^4\left(5^{2017}-5\right)+5^5=5^4\left(5^{2017}-5\right)+3125=\)

\(=5^4\left(5^{2017}-5\right)+2017+1108\)

Ta có

\(5^4\left(5^{2017}-5\right)+2017⋮2017\)

\(\Rightarrow5^{2021}\equiv1108\) (mod 2017)

\(x\left(x-8\right)=9\)

\(\Leftrightarrow x^2-8x-9=0\)

\(\Leftrightarrow x\left(x-9\right)+\left(x-9\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=9\) hoặc \(x=-1\)

\(x\left(x+14\right)=15\)

\(\Leftrightarrow x^2+14x-15=0\)

\(\Leftrightarrow x\left(x+15\right)-\left(x+15\right)=0\)

\(\Leftrightarrow\left(x+15\right)\left(x-1\right)=0\)

\(\Leftrightarrow x=-15\) hoặc \(x=1\)

\(4x\left(x-3\right)=7\)

\(\Leftrightarrow4x^2-12x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\) hoặc \(x=\dfrac{7}{2}\)

\(9x\left(x-4\right)=13\)

\(\Leftrightarrow9x^2-36x-13=0\)

\(\Leftrightarrow3x\left(3x+1\right)-13\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(3x-13\right)=0\)

\(\Leftrightarrow x=-\dfrac{1}{3}\) hoặc \(x=\dfrac{13}{3}\)

\(\left(x+6\right)^2-x\left(x-3\right)=81\)

\(\Leftrightarrow x^2+12x+36-x^2+3x=81\)

\(\Leftrightarrow15x=45\)

\(\Leftrightarrow x=3\)

\(\left(2x+3\right)^2-x\left(x-3\right)=9\)

\(\Leftrightarrow4x^2+12x+9-x^2+3x=9\)

\(\Leftrightarrow3x^2+15x+9=9\)

\(\Leftrightarrow3x\left(x+5\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(x=-5\)

\(\left(x-7\right)^2-x\left(x+4\right)=121\)

\(\Leftrightarrow x^2-14x+49-x^2-4x=121\)

\(\Leftrightarrow-18x=72\)

\(\Leftrightarrow x=-4\)

\(\left(3x-2\right)^2-9x\left(x-4\right)=100\)

\(\Leftrightarrow9x^2-12x+4-9x^2+36x=100\)

\(\Leftrightarrow24x=96\)

\(\Leftrightarrow x=4\)

\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^3=\left(\dfrac{1}{2}a\right)^3-3.\left(\dfrac{1}{2}a\right)^2.\dfrac{2}{3}b+3.\dfrac{1}{2}a.\left(\dfrac{2}{3}b\right)^2-\left(\dfrac{2}{3}b\right)^3\)

                        \(=\dfrac{1}{8}a^3-\dfrac{1}{2}a^2b+\dfrac{2}{3}ab^2-\dfrac{8}{27}b^3\)

\(\left(\dfrac{1}{2}a-\dfrac{2}{3}b\right)^2\)

\(=\left(\dfrac{1}{2}a\right)^2-2.\left(\dfrac{1}{2}a\right).\left(\dfrac{2}{3}b\right)+\left(\dfrac{2}{3}b\right)^2\)

\(=\dfrac{1}{4}a^2-\dfrac{2}{3}ab+\dfrac{4}{9}b^2\)