\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)

\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)

TUI ĐANG GẤP CHO TÔI HỎI BÀI NÀY LỚP 2 NHA\\\\

AN CÓ 180 CÁI KẸO.BÌNH CÓ 160. HỎI BÌNH CÓ MẤY CÁI KẸO

a) Ta có: \(2.4.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)

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

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

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

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

\(=3^{32}-1\)

$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$

$=\dfrac{3^{32}-1}{2}$

c;=(50-49)(50+49)+(48-47)(48+47)+.............+(2+1)(2-1)

=50+49+48+............+1

=(50+1)50=2550:2=1275

d;=(2^4-1)(2^4+1)(2^8+1)(2^16+1)

=(2^8-1)(2^8+1)(2^16+1)

=(2^16-1)(2^16+1)

=2^32-1

e;=(3-1)(3+1)(3^2+1)...........(3^16+1)

=(3^2-1)(3^2+1)..............(3^16+1)

=(3^16-1)(3^16+1)=3^32-1

tu tinh ket qua luy thua tao khong thua hoi dau

1/

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

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

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

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

\(=\frac{3^{16}-1-3^{16}}{4}=\frac{-1}{4}\)

2/

a, (x-5)²-(x+3)²=1

<=>(x-5+x+3)(x-5-x-3)=1

<=>-16.(x-1)=1

<=>x-1=-1/16

<=>x=15/16

b, (2x-1)²-(2x-3)²=4

<=>(2x-1+2x-3)(2x-1-2x+3)=4

<=>-8(x-1)=4

<=>x-1=-1/2

<=>x=1/2

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

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

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

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

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

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

=\(2^{64}-1\)

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

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

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

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

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

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

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

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

mk cảm ơn ah