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106104.7037 đây là chuẩn còn làm tròn là 106105 nha

\(57296:54=\dfrac{28648}{27}\)

\(3^{x-1}+3^x+3^{x+1}=39\)

\(3^{x-1}+3^{x-1}.3+9.3^{x-1}=39\)

\(13.3^{x-1}=39\)

\(3^{x-1}=39:13=3\)

\(x-1=1\)

\(x=2\)

Sửa đề: 3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39

3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39

3ˣ⁻¹.(1 + 3 + 3²) = 39

3ˣ⁻¹ . 13 = 39

3ˣ⁻¹ = 39 : 13

3ˣ⁻¹ = 3

x - 1 = 1

x = 1 + 1

x = 2

\(3^{n+1}-2.3^n+2^{n+5}-7.2^n\)

\(=3^n\left(3-2\right)+2^n\left(2^5-7\right)\)

\(=3^n+2^n.25\)

Vì \(3^n⋮̸25\); \(25.2^n⋮25\)

=> \(3^n+2^n.25⋮̸25\)

=> \(3^{n+1}-2.3^n+2^{n+5}-7.2^n⋮̸25\)

\(313^5.299-313^6.36\)

\(=313^5.299-313^636\)

\(=313^5\left(299-313.36\right)\)

Ta có:

Ta có: \(299\equiv5\left(mod7\right)\)

       \(313\equiv5\left(mod7\right)\)

       \(36\equiv1\left(mod7\right)\)

=> \(299-313.36\equiv5-5.1=0\left(mod7\right)\)

=> \(299-313.36⋮7\)

=> \(313^5.299-313^6.36⋮7\)

\(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)

\(=\dfrac{1}{3}\left(-1+\dfrac{1}{3}\right)+\dfrac{1}{3^3}\left(-1+\dfrac{1}{3}\right)+...+\dfrac{1}{3^{99}}\left(-1+\dfrac{1}{3}\right)\)

\(=\dfrac{-2}{3}\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)

\(9B-B=3-\dfrac{1}{3^{99}}\)

\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)

\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)

Ta có:

\(x+y=1\Rightarrow3xy=3xy\left(x+y\right)\)

\(x^3+3xy+y^3\)

\(=x^3+3xy\left(x+y\right)+y^3\)

\(=\left(x+y\right)^3=1\)

\(D=-\dfrac{4}{5}+\dfrac{4}{5^2}-\dfrac{4}{5^3}+...+\dfrac{4}{5^{200}}\)

\(\Rightarrow D=4\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)

\(5D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}\right)\)

\(\Rightarrow5D+D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)

\(\Rightarrow6D=4\cdot\left(\dfrac{1}{5^{200}}-1\right)\)

\(\Rightarrow D=\dfrac{2}{3}\cdot\left(\dfrac{1}{5^{200}}-1\right)\)

oki

\(B=7-7^4+7^7-7^{10}+...+7^{295}-7^{298}+7^{301}\)

\(=7\left(1-7^3\right)+7^7\left(1-7^3\right)+...+7^{295}\left(1-7^3\right)+7^{301}\)

\(=\left(1-7^3\right)\left(7+7^7+...+7^{295}\right)+7^{301}\)

\(=\left(1-7^3\right)\left(\dfrac{7^{296}-7}{6}\right)+7^{301}\)

\(=-57\left(7^{296}-7\right)+7^{301}\)

\(\left(-0,25\right)^4\cdot4^4\)

\(=\left(-\dfrac{1}{4}\right)^4\cdot4^4\)

\(=\left(-\dfrac{1}{4}\cdot4\right)^4\)

\(=\left(-\dfrac{4}{4}\right)^4\)

\(=\left(-1\right)^4\)

\(=1\)

(-0,25)⁴.4⁴

= (-0,25 . 4)⁴

= (-1)⁴

= 1