a: \(P=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^{32}-1\right)\left(3^{32}+1\right)}{2}=\dfrac{3^{64}-1}{2}\)

b: \(Q=\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)}{5^2-1}\)

\(=\dfrac{\left(5^{32}-1\right)\left(5^{32}+1\right)}{24}=\dfrac{5^{64}-1}{24}\)

a: \(\left|0,5x-2\right|-\left|x+\dfrac{2}{3}\right|=0\)

=>\(\left|\dfrac{1}{2}x-2\right|=\left|x+\dfrac{2}{3}\right|\)

=>\(\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+\dfrac{2}{3}\\\dfrac{1}{2}x-2=-x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x=2+\dfrac{2}{3}=\dfrac{8}{3}\\\dfrac{3}{2}x=-\dfrac{2}{3}+2=\dfrac{4}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{8}{3}:\dfrac{-1}{2}=\dfrac{8}{3}\cdot\left(-2\right)=-\dfrac{16}{3}\\x=\dfrac{4}{3}:\dfrac{3}{2}=\dfrac{8}{9}\end{matrix}\right.\)

b: 

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

=>\(\left|x+1\right|=2x-\dfrac{1}{4}\)

=>\(\left\{{}\begin{matrix}2x-\dfrac{1}{4}>=0\\\left(2x-\dfrac{1}{4}\right)^2=\left(x+1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(2x-\dfrac{1}{4}-x-1\right)\left(2x-\dfrac{1}{4}+x+1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(x-\dfrac{5}{4}\right)\left(3x+\dfrac{3}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{4}\)

c: \(3x-\left|x+15\right|=\dfrac{5}{4}\)

=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)

=>\(\left\{{}\begin{matrix}3x-\dfrac{5}{4}>=0\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16,25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=8,125\)

d: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

=>\(\left|3x+\dfrac{5}{4}\right|=\dfrac{3}{2}-\dfrac{1}{4}=\dfrac{5}{4}\)

=>\(\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

e: \(\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\)

=>\(\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=-3x+\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-3x=-\dfrac{1}{2}+1\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

f: \(\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)

=>\(\left[{}\begin{matrix}2x-1=x+\dfrac{1}{3}\\2x-1=-x-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=\dfrac{1}{3}+1\\2x+x=-\dfrac{1}{3}+1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{4}{3}\\3x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{2}{9}\end{matrix}\right.\)

Xét tổng: S = 1 + 2 + 3 + ... + x 

Số lượng số hạng: (x - 1) :  1 + 1 = x (số hạng)

`=>S=((x+1)*x)/2` 

\(=>\dfrac{\left(x+1\right)x}{2}=820\\ =>x\left(x+1\right)=1640\\ =>x^2+x-1640=0\\ =>\left(x^2-40x\right)+\left(41x-1640\right)=0\\ =>x\left(x-40\right)+41\left(x-40\right)=0\\ =>\left(x+41\right)\left(x-40\right)=0\\ =>\left[{}\begin{matrix}x=40\\x=-41\end{matrix}\right.\)

Mà x > 0 => x = 40 

= 0,5

=50

Chiều cao của thửa ruộng là:

\(\dfrac{5}{6}\times120=100\left(m\right)\)

Diện tích của thửa ruộng là:

\(120\times100=12000\left(m^2\right)\)

Khối lượng lúa thu hoạch được là:
\(12000:500\times1250=30000\left(kg\right)\)

Số tiền thu được là:

\(30000\times17600=528000000\left(đ\right)\)

ĐS: .. 

Chiều cao thửa ruộng đó là:

\(120\times\dfrac{5}{6}=100\left(m\right)\)

Diện tích thửa ruộng là:

\(120\times100=12000\left(m^2\right)\)

Số lúa thu hoạch được trên thửa ruộng đó là:

\(1250:500\times12000=30000\left(kg\right)\)

Số tiền thu được từ thửa ruộng là:

\(30000\times17600=528000000\) (đồng)

\(a.\left(x^2+5x+6\right)\left(x^2-15x+56\right)-144\\ =\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)-144\\ =\left[\left(x+2\right)\left(x-7\right)\right]\left[\left(x+3\right)\left(x-8\right)\right]-144\\ =\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\\ =\left(x^2-5x-19+5\right)\left(x^2-5x-19-5\right)-144\\ =\left(x^2-5x-19\right)^2-5^2-144\\ =\left(x^2-5x-19\right)^2-169\\ =\left(x^2-5x+19\right)^2-13^2\\ =\left(x^2-5x-19-13\right)\left(x^2-5x-19+13\right)\\ =\left(x^2-5x-32\right)\left(x^2-5x-6\right)\\ =\left(x^2-5x-32\right)\left(x+1\right)\left(x-6\right)\) 

\(b.\left(x^2-11x+28\right)\left(x^2-7x+10\right)-72\\ =\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-2\right)-72\\ =\left[\left(x-4\right)\left(x-5\right)\right]\left[\left(x-7\right)\left(x-2\right)\right]-72\\ =\left(x^2-9x+20\right)\left(x^2-9x+14\right)-72\\ =\left(x^2-9x+17+3\right)\left(x^2-9x+17-3\right)-72\\ =\left(x^2-9x+17\right)^2-3^2-72\\ =\left(x^2-9x+17\right)^2-81\\ =\left(x^2-9x+17\right)^2-9^2\\ =\left(x^2-9x+17-9\right)\left(x^2-9x+17+9\right)\\ =\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)

8

2³.6 - 72 : 3²

= 8.6 - 72 : 9

= 48 - 8

= 40

a)

A = -4x² - 12x

= -4(x² + 3x)

b)

B = 3 - 4x - x²

= -(x² + 4x - 3)

= -(x² + 4x + 4 - 7)

= -(x + 2)² + 7

Do (x + 2)² ≥ 0

⇒ -(x + 2)² ≤ 0

⇒ -(x + 2)² + 7 ≤ 7

Vậy maxB = 7 khi x = -2

\(x^{64}+x^{32}+1\\ =x^{64}+2x^{32}+1+x^{32}-2x^{32}\\ =\left[\left(x^{32}\right)^2+2\cdot x^{32}\cdot1+1^2\right]-x^{32}\\ =\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\\ =\left(x^{32}-x^{16}+1\right)\left(x^{32}+x^{16}+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{32}+2x^{16}+1\right)+x^{16}-2x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left[\left(x^{16}+1\right)^2-x^{16}\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^{16}+x^8+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^{16}+2x^8+1\right)-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left[\left(x^8+1\right)^2-x^8\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^8+x^4+1\right)\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^8+2x^4+1\right)-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left[\left(x^4+1\right)^2-x^4\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)

\(=\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^4+2x^2+1\right)-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]\\ =\left(x^{32}-x^{16}+1\right)\left(x^{16}-x^8+1\right)\left(x^8-x^4+1\right)\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)

\(x^{64}+x^{32}+1\)

\(=x^{64}+2x^{32}-x^{32}+1\)

\(=\left(x^{64}+2^{32}+1\right)-x^{32}\)

\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)

\(=\left(x^{32}+1-x^{16}\right)\left(x^{32}+1+x^{16}\right)\)