\(\text{Xét công thức tổng quát }:x^4+\frac{1}{4}=\left(x^4+2.x^2.\frac{1}{2}+\frac{1}{4}\right)-x^2\)

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

Áp dụng vào B ta đc:

\(B=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}\)

\(=\frac{\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)...\left(11^2-11+\frac{1}{2}\right)\left(11^2+11+\frac{1}{2}\right)}{\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)...\left(12^2-12+\frac{1}{2}\right)\left(12^2+12+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(122+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(156+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}\left(122+\frac{1}{2}\right)}{\left(132+\frac{1}{2}\right)\left(156+\frac{1}{2}\right)}=\frac{49}{16589}\)

ko biết có đúng ko!! hình như còn 1 cách là nhân 1 đa thức với 16 nữa thì phải lâu ko động đến bạn thử xem đc ko nhé

a)\({\left( { - 2} \right)^2}.{\left( { - 2} \right)^3} = {\left( { - 2} \right)^{2 + 3}} = {\left( { - 2} \right)^5}\);

b)\({\left( { - 0,25} \right)^7}:{\left( { - 0,25} \right)^5} = {\left( { - 0,25} \right)^{7 - 5}} = {\left( { - 0,25} \right)^2} = {\left( {0,25} \right)^2}\);

c)\({\left( {\frac{3}{4}} \right)^4}.{\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^{4 + 3}} = {\left( {\frac{3}{4}} \right)^7}.\)

\(x=\left(1-\dfrac{1}{\sqrt{4}}\right).\left(1-\dfrac{1}{\sqrt{16}}\right).\left(1-\dfrac{1}{\sqrt{36}}\right).\left(1-\dfrac{1}{\sqrt{64}}\right).\left(1-\dfrac{1}{\sqrt{100}}\right)\)

\(x=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{8}\right).\left(1-\dfrac{1}{10}\right)\)

\(x=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)

\(x=\dfrac{63}{256}\)

và \(y=\sqrt{20+0,25}\)

\(y=\sqrt{20,25}\)

\(y=4,5\)

Do 4,5 > \(\dfrac{63}{256}\)

=> x<y

5.[3/4-2/3x]-13/4=-2{(5/4)²-5/2.1/4+(1/16)²

5.[3/4-2/3x]-13/4=2{25/16-5/8+1/256}

........................=2{400/256-160/250+1/256}

........................=2.241/256

........................=241/128

5.[3/4-2/3x]=241/128+13/4

5.[3/4-2/3x]=241/128+416/128

[3/4-2/3x]=657/128:5

[3/4-2/3x]=657/640

vậy 3/4-2/3x=657/640 và 3/4-2/3x=-657/640

th1:3/4-2/3x va th2:3/4-2/3x=-657/640

còn lại bạn chỉ cần tính kết quả mà dấu[..] là giá tri tuyệt đối đó

\(x:\left(0,25\right)^4=\left(0,5\right)^2\)

\(\Rightarrow x:\left(0,25\right)^4=0,25\)

\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)

\(\Rightarrow x=\left(0,25\right)^5\)

\(\Rightarrow x=\frac{1}{1024}\)

Vậy \(x=\frac{1}{1024}.\)

Chúc bạn học tốt!

\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)

\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)

a) \(...=0,25+1500+\left(0,5\right)^2=0,25+0,25=1500=1500,5\)

b) \(...=2,7-4,4+5,6-7,3=2,7+5,6-4,4-7,3=8,3-11,7=-3,4\)

c) \(...=-5,44+5+0,44=-5,44+0,44+5=-5+5=0\)

d) \(...=6,72+5,27-0,72-1,27=6,72-0,72+5,27-1,27=6+4=10\)

a) 1500,5

b) -3,4

c) 0

d) 10