a.

\(2^x.4=16\)

\(2^x=16:4\)

\(2^x=4\)

\(2^x=2^2\)

\(x=2\)

b.

\(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(x=4\)

c.

\(64.4^x=16^8\)

\(4^3.4^x=\left(4^2\right)^8\)

\(4^{x+3}=4^{2.8}\)

\(4^{x+3}=4^{16}\)

\(x+3=16\)

\(x=13\)

d.

\(2^x.16^x=1024\)

\(2^x.\left(2^4\right)^x=2^{10}\)

\(2^x.2^{4x}=2^{10}\)

\(2^{x+4x}=2^{10}\)

\(2^{5x}=2^{10}\)

\(5x=10\)

\(x=2\)

\(\dfrac{x-5}{3}=\dfrac{x+7}{2}\\ =>2\left(x-5\right)=3\left(x+7\right)\\ =>2x-10=3x+21\\ =>3x-2x=-10-21\\ =>x=-31\)

Vậy: ... 

\(\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{97\cdot95}+...+\dfrac{1}{5\cdot3}+\dfrac{1}{3\cdot1}\right)\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\\ =\dfrac{1}{99\cdot97}-\left(-\dfrac{1}{97}+1\right)\\ =\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-1\\ =\dfrac{1}{99}-1\\ =-\dfrac{98}{99}\)

\(\dfrac{11}{48}=\dfrac{11.3}{48.3}=\dfrac{33}{144}\)

\(\dfrac{17}{36}=\dfrac{17.4}{36.4}=\dfrac{68}{144}\)

Do \(68>33\Rightarrow\dfrac{68}{144}>\dfrac{33}{144}\Rightarrow\dfrac{17}{36}>\dfrac{11}{48}\)

Có thể lập được 6 số là: 1350, 1530, 3150, 3510, 5130, 5310

Có thể lập được 6 số là: 1350, 1530, 3150, 3510, 5130, 5310

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Giúp mình với ạ cầu xin mọi người luôn đó 

\(65180=60000+5000+100+80=6.10^4+5.10^3+1.10^2+8.10^1\)

\(101010=100000+1000+10=1.10^5+1.10^3+1.10^1\)

\(\overline{ab0cd}=\overline{a0000}+\overline{b000}+\overline{c0}+d=a.10^4+b.10^3+c.10^1+d.10^0\)