\(25^{2x}:5^x=125^2\)

\(\Rightarrow5^{4x}:5^x=\left(5^3\right)^2\)

\(\Rightarrow5^{4x-x}=5^6\)

\(\Rightarrow5^{3x}=5^6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

\(1-2+3-4+5-6+...+2019-2020+2021\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021\)

\(=-1-1-1-..-1+2021\)

\(=-1\cdot1010+2021\)

\(=-1010+2021\)

\(=-1011\)

\(1000=2^3\cdot5^3\)

\(7200=2^5\cdot3^2\cdot5^2\)

\(810=2\cdot3^4\cdot5\)

\(8000=2^6\cdot5^3\)

a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x^2-x+1>0\forall x\)

2n + 19 chia hết cho 2n + 5 

⇒ 2n + 5 + 14 chia hết cho 2n + 5 

⇒ 14 chia hết cho 2n + 5 

⇒ 2n + 5 ϵ Ư(14) 

Mà n nguyên nên 2n + 5 ϵ { 1 ; -1 ; 7; -7)

Ta có bảng sau:

Vậy: n ϵ {-2 ; -3; 1; -6}

\(\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(=\left(2^3\cdot3^8+3^6\cdot5\cdot3^2\right):\left[9^2\cdot\left(10-1\right)\right]\)

\(=\left(2^3\cdot3^8+3^8\cdot5\right):\left(9^2\cdot9\right)\)

\(=\left[3^8\cdot\left(2^3+5\right)\right]:9^3\)

\(=3^8\cdot13:3^6\)

\(=3^2\cdot13\)

\(=117\)

\(14-x+\left(-10\right)=-\left(-9\right)+\left(+15\right)\)

\(\Rightarrow14-x-10=9+15\)

\(\Rightarrow4-x=24\)

\(\Rightarrow x=4-24\)

\(\Rightarrow x=-20\)

\(100:\left\{250:\left[450-\left(4\cdot5^3-2^2\cdot25\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(4\cdot125-4\cdot25\right)\right]\right\}\)

\(=100:\left\{250:\left[450-\left(500-100\right)\right]\right\}\)

\(=100:\left[250:\left(450-400\right)\right]\)

\(=100:\left(250:50\right)\)

\(=100:5\)

\(=20\)

\(a^2+16\cdot a\)

\(=a\cdot\left(a-16\right)\)

Diện tích ban đầu của tấm bìa:

\(350:\dfrac{1}{4}=1400\left(dm^2\right)\)

Đổi: \(1400\left(dm^2\right)=14\left(m^2\right)\)

Đáp số: \(14\left(m^2\right)\)