Căn 2 + Căn 8 + Căn 50 mới bằng 8 căn 2

Cũng tùi theo may mắn

cũng tùi theo may mắn

Thì bạn chỉ cần trả lời đúng,trình bày đẹp,dễ hiểu thôi và chăm chỉ trả lời

Ta có: $A=\left(\genfrac{}{}{0ex}{}{1}{4\cdot 9}+\genfrac{}{}{0ex}{}{1}{9\cdot 14}+\genfrac{}{}{0ex}{}{1}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{1}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{5}{4\cdot 9}+\genfrac{}{}{0ex}{}{5}{9\cdot 14}+\genfrac{}{}{0ex}{}{5}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{5}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{9}+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{14}+\genfrac{}{}{0ex}{}{1}{14}-\genfrac{}{}{0ex}{}{1}{19}+...+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{49-4}{4\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{45}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{-623}{89}=-\genfrac{}{}{0ex}{}{9}{28}$
 

\(C=1+4^1+4^2+3^3+...+4^{31}\)

\(4C=4\left(1+4^1+4^2+4^3+...+4^{31}\right)\)

\(4C=4+4^2+4^3+4^4+...+4^{32}\)

\(4C-C=\left(4+4^2+4^3+4^4+...+4^{32}\right)-\left(1+4^1+4^2+4^3+...+4^{31}\right)\)

\(3C=4^{32}-1\)

\(C=\dfrac{4^{32}-1}{3}\)

a)

Ta có:

\(24=2^3.3\)

\(30=2.3.5\)

\(\RightarrowƯCLN\left(24;30\right)=2.3=6\)

\(\RightarrowƯC\left(24;30\right)=Ư\left(6\right)=\left\{1;2;3;6\right\}\)

b)

Ta có:

\(42=2.3.7\)

\(98=2.7^2\)

\(\RightarrowƯCLN\left(42;98\right)=2.7=14\)

\(\RightarrowƯC\left(42;98\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\)

c)

Ta có:

\(180=2^2.3^2.5\)

\(234=2.3^2.13\)
\(\RightarrowƯCLN\left(180;234\right)=2.3^2=18\)

\(\RightarrowƯC\left(180;234\right)=Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

\(2x+3x+5x+...+17x+19x\)

\(x.\left(2+3+5+...+17+19\right)=1+2+...+9\)

\(x.101=45\)

\(\Rightarrow x=45:101\)

\(\Rightarrow x=\dfrac{45}{101}\)

\(53.x+2-25=3100\)

\(53.x+2=3100+25\)

\(53.x+2=3125\)

\(53.x=3125-2\)

\(53.x=3123\)

\(x=3123:53\)

\(x=\dfrac{3123}{53}\)