\(a,\left(2x-3\right)n-2n\left(n+2\right)\)

\(=n\left(2x-3-2n-4\right)\)

\(=-7n\)

Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM

\(b,n\left(2n-3\right)-2n\left(n+1\right)\)

\(=n\left(2n-3-2n-2\right)\)

\(=-5n⋮5\) (ĐPCM)

Rút gọn

\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)

\(=-76\)

\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)

\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)

\(=9\)

\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)

\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)

= -3

a ) Ta có : \(f\left(x\right)=4x^2-4x+3=4x^2-4x+1+2\)

\(=\left(2x-1\right)^2+2\ge2>0\forall x,x\in R\)

b ) Ta có : \(g\left(x\right)=2x-x^2-7=-x^2+2x-7\)

\(=-x^2+2x-1-8\)

\(=-\left(x^2-2x+1\right)-8\)

\(=-\left(x-1\right)^2\le-8< 0\forall x,x\in R\)

Lời giải:

Biến đổi: \(q(x)=9.81^x+15.25^x+2.8^x+8.64^x\)

Lại có:

\(\left\{\begin{matrix} 81\equiv 13\pmod {17}\rightarrow 81^k\equiv 13^k\pmod {17}\\ 25\equiv 8\pmod {17}\rightarrow 25^k\equiv 8^k\pmod {17}\\ 64\equiv 13\pmod {17}\rightarrow 64^k\equiv 13^k\pmod {17}\end{matrix}\right.\)

Do đó, \(q(x)\equiv 9.13^k+15.8^k+2.8^k+8.13^k\pmod {17}\)

\(\Leftrightarrow q(x)\equiv 17.13^k+17.8^k\equiv 0\pmod {17}\)

\(\Leftrightarrow q(x)\vdots 17\) (đpcm)

yugyuf

\(\text{f(x)}\)\(\text{>0}\)\(\text{⇔}\)\(\text{2x}\)²\(\text{-3x+1}\)\(>0\)⇔\(\left\{{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\)

⇒x∈(−∞;\(\dfrac{1}{2}\))∪(1;+∞)

Vì đây là 7 số liên tiếp

nên A chia hết cho 7!

=>A chia hết cho 210