15(4^2+1)(4^4+1)...(4^64+1)

- Vậy 

\(15\cdot\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)....\left(4^{64}+1\right)\)

\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)....\left(4^{64}+1\right)\)

\(=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)

\(=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(=\left(4^{64}-1\right)\left(4^{64}+1\right)\)

\(=4^{128}+1\)

lỗi nên đăng lại nhe

15(4^2+1)(4^4+1)...(4^64+1)

- Vậy 

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

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

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

Ta có (4² - 1)(4² + 1) = 4⁴ - 1

Ta có 15A = (4² - 1)(4² + 1)(4⁴ + 1)(4⁸ + 1)(4¹⁶ + 1)(4³² + 1) - 4⁶⁴ = 4⁶⁴ - 1 - 4⁶⁴  = -1

=> A = \(\frac{-1}{15}\)

Ghi lại cái đề cho rõ hơn đi t giải cho

Cái đề đọc không ra.Ghi lại cái đề cho rõ hơn đi t giải cho

a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)

b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)

c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)

d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)

e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)

mn giúp e vs ạT^T

\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)

\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)

3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

=(5^8-1)(5^8+1)(5^16+1)

=(5^16-1)(5^16+1)

=5^32-1

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(C=5^{32}-1\)

4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)

\(D=4^{128}-1\)

5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)

....

\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)

\(E=5^{512}-1\)

Bài 2: 

Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)

Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)

                                 = 100000 ( do 1⁵ cũng sẽ =1 nên không viết mũ 5 cũng chả sao)