giúp mình với mình đang cần gấp

\(\Rightarrow4^x=4^4:4^3=4^1\\ \Rightarrow x=1\)

⇒ 4^x . 4³ = 4⁴

⇒ x + 3 = 4

⇒ x       = 1

ban viet lai de bai duoc ko minh ko hieu neu viet so mu thi an shift roi an so 7 la mu vi du 3^2

đề :    CMR :  1+4+4^2+4^3+..........+4^58+4^59 chia hết cho 21,85

mk chỉ viết đề thôi chứ bn thông cảm mk chưa hok dạng này

Ta thấy :

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

..............

\(\dfrac{1}{99^2}>\dfrac{1}{99.100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)

Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Dựa vào công thức ta có :

\(\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

.......................

\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{100}\)

\(\Rightarrow K>\dfrac{6}{25}>\dfrac{1}{5}\Rightarrow dpcm\) (1)

Ta có :

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

................

\(\dfrac{1}{99^2}< \dfrac{1}{98.99}\)

Dựa vào công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\) ta có :

\(K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{98.99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.......+\dfrac{1}{98}-\dfrac{1}{99}\)

\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{99}\)

Vậy \(K< \dfrac{32}{99}< \dfrac{1}{3}\Rightarrow dpcm\) (2)

(1) ; (2) \(\Rightarrow\) \(\dfrac{1}{5}< K< \dfrac{1}{3}\)

Ai thấy đúng thì ủng hộ nha !!!

Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)

=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)

=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)

a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{5}{7}\times1\)

\(=\dfrac{5}{7}\)

b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)

\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)

\(=1+0\)

\(=1\)

c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)

\(=\dfrac{5}{7}+\dfrac{2}{7}\)

\(=1\)

d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)

\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)

\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)

\(=\dfrac{1}{2}+1\)

\(=\dfrac{3}{2}\)

e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)

\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)

\(=\dfrac{4}{5}+\dfrac{12}{10}\)

\(=\dfrac{4}{5}+\dfrac{6}{5}\)

\(=\dfrac{10}{5}\)

\(=2\)

g) \(362\times728+326\times272\)

\(=326\times\left(728+272\right)\)

\(=326\times1000\)

\(=326000\) 

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

Hướng dẫn giải:

a) 3 x (20 – 5)

Cách 1:

3 x (20 – 5) = 3 x 15 = 45

Cách 2:

3 x (20 – 5) = 3 x 20 – 3 x 5 = 60 – 15 = 45

b) 20 x (40 – 1)

Cách 1:

20 x (40 – 1) = 20 x 39 = 780

Cách 2:

20 x (40 – 1) = 20 x 40 – 20 x 1 = 800 – 20 = 780