a)=648

c) gọi biểu thức là S = 2 + 2\(^2+2^3+.....+2^{50}\)

2S=2\(^2+2^3+2^4+......+2^{50}+2^{51}\)

\(2S-S=S=2^{51}-2\)

b) \(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{10}}\)

= \(2+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^9}\)

2S-S=S=(\(2+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^9}\))-( \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\))

bạn tự tìm S nhé

mink làm được như thế đó, phần a mink không muốn nhấn mỏi tay bạn ạ, đừng nghĩ mink ko biết làm nha

a) TICK CHO MÌNH VS\(\dfrac{5.27^3.4^5}{6^5}:\left(\dfrac{5^5.2^4}{10^4}.\dfrac{2^6.3^4}{6^4}\right)=\dfrac{100776960}{7776}:\left(\dfrac{50000}{10000}.\dfrac{5184}{1296}\right)=12960:\left(5.4\right)=12960:20=648\)

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\) 

\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) 

\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\) 

Rút gọn vế trái ta có :

\(2^5.2.2.^5=2^n\)

\(\Rightarrow2^{12}=2^n\) 

\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) ) 

Vậy n =12 

=>\(\dfrac{4^5\left(1+1+1+1\right)}{3^5\left(1+1+1\right)}.\dfrac{6^5\left(1+1+1+1+1+1\right)}{2^5\left(1+1\right)}=2^n\)

=>\(\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) =>\(\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=2^n\)

=>\(\left(\dfrac{4.6}{3.2}\right)^6=2^n\) =>\(4^6=2^n\) =>\(2^{12}=2^n\) =>n=12.

a) \(\dfrac{1}{4}+\left(-\dfrac{3}{8}\right)=\dfrac{2}{8}+\left(-\dfrac{3}{8}\right)=-\dfrac{1}{8}\)
b)\(\dfrac{2^{10}.3^4}{3^5.2^8}=\dfrac{2^8.2^2.3^4}{3^4.3.2^8}=\dfrac{2^2}{3}=\dfrac{4}{3}\)
c) \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{9}}=\dfrac{5}{10}.10-\dfrac{1}{3}=5-\dfrac{1}{3}=\dfrac{15}{3}-\dfrac{1}{3}=\dfrac{14}{3}\)
d) \(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+5\dfrac{4}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\dfrac{41}{9}:\left(-\dfrac{5}{7}\right) +\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
\(=\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
\(=10:\left(-\dfrac{5}{7}\right)\)
\(=10.\left(-\dfrac{7}{5}\right)\)
\(=-14\)

a: \(\dfrac{5^5}{5^x}=5^{18}\)

=>5-x=18

hay x=-13

b: \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)

=>4-x=50

hay x=-46

c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)

=>2x-3=49

=>2x=52

hay x=26

d: \(\dfrac{2^3}{2^x}=4^5\)

\(\Leftrightarrow2^{3-x}=2^{10}\)

=>3-x=10

hay x=-7

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)

\(\Leftrightarrow5^x=5^6\)

hay x=6

f: \(7\cdot2^x=2^9+5\cdot2^8\)

\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)

\(\Leftrightarrow2^x=2^8\)

hay x=8

Câu hỏi của Lê Khánh Nhi - Toán lớp 7 - Học toán với OnlineMath sửa n thành x cho sửa cho nó thành lũy thừa luôn

1; 5.2² + (\(x\) + 3) = 5²

   5.4  +  (\(x\) + 3) = 25

    20 + (\(x\) + 3) = 25

             \(x\) + 3 = 25 - 20

             \(x+3\) = 5

             \(x\)       = 5  - 3

            \(x\)        = 2

            Vậy \(x=2\)

2; 2³ + (\(x\) - 3²) = 5³ - 4³

   8 +  (\(x\) - 9)    = 125 - 64

   8 + (\(x\) - 9) = 61

         \(x\) - 9 = 61 - 8

         \(x\) - 9 = 53

        \(x\)        = 53  + 9

        \(x\)       = 62

        Vậy \(x\) = 62

3; 3.2 + (\(x\) + 5²) = 10²

       6 + (\(x\) + 25) = 100

              \(x\) + 25 = 100 - 6

              \(x\) + 25  = 94

               \(x\)          = 94 - 25

               \(x\)          = 69

Vậy \(x\) = 69

p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)