a: x-56:4=16

nên x-14=16

hay x=30

b: \(101+\left(36-4x\right)=105\)

\(\Leftrightarrow36-4x=4\)

hay x=8

Vì a: x-56:4=16

    =>x-14=16

   Hay:x=30

Vì b: 

   Hay:x=8

         Vậy x={30:8}

 Nếu sai thì thui nhá,đừng chửi mình!

Giải :

72^45-72^44 và 72^49-72^33 

Khi trừ luỹ thừa cùng cơ số, ta giữ nguyên cơ số và trừ số mũ. 

=72^45-44 và 72^49-33

=72^1 và 72^16

Do 72^1 < 72^16 nên 72^45 - 72^44 < 72^49 - 72^33

Vậy 72^45 - 72^44 < 72^49 - 72^33

Ý mình sai rồi, xin lổi nha bạn.

Giải :

72^45-72^44 và 72^49-72^33

= 72^(45-44) và 72^(49-33)

= 72^1 < 72^16

Do 72^1 < 72^16 nên 72^45-72^44 < 72^49-72^33

Vậy 72^45-72^44 < 72^49-72^33

a: 7x+58=100

nên 7x=42

hay x=6

c: x-56:x=16

nên x-14=16

hay x=30

c)x - 56 : 4  = 16

x - 56       = 16 : 4 

x- 56        = 4

x               =4 + 56

x               = 60 

d)101 + (36 - 4x) = 105 

          (36- 4x ) = 105 - 101 

          36 - 4x = 4

                4x = 36 - 4 

                4x = 32

                  x = 32:4

                  x = 8

a) 58 + 7x = 100

            7x = 100 - 58

            7x = 42

              x = 42 : 7

              x = 6

b) (x - 12) : 12 = 12

              x - 12 = 12 . 12

             x - 12 = 144

                    x = 144 + 12

                    x = 156

\(\dfrac{4x}{5}+\dfrac{4x}{45}+\dfrac{4x}{117}+...+\dfrac{4x}{1221}=\dfrac{72}{37}\)

\(\dfrac{4x}{1.5}+\dfrac{4x}{5.9}+\dfrac{4x}{9.13}+...+\dfrac{4x}{33.37}=\dfrac{72}{37}\)

\(\Leftrightarrow4x\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{33.37}\right)=\dfrac{72}{37}\)

\(\Leftrightarrow4x\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)=\dfrac{72}{37}\)

\(\Leftrightarrow4x\left(1-\dfrac{1}{37}\right)=\dfrac{72}{37}\)

\(\Leftrightarrow4x.\dfrac{36}{37}=\dfrac{72}{37}\)

\(4x=\dfrac{72}{37}:\dfrac{36}{37}\)

\(4x=2\)

\(x=2:4\)

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

\(\dfrac{4x}{5}+\dfrac{4x}{45}+\dfrac{4x}{117}+...+\dfrac{4x}{1221}=\dfrac{72}{37}\)

\(\Rightarrow\) \(\dfrac{4x}{1\cdot5}+\dfrac{4x}{5\cdot9}+\dfrac{4x}{9\cdot13}+...+\dfrac{4x}{33\cdot37}=\dfrac{72}{37}\)

\(\Rightarrow\) \(4x\left[\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{33\cdot37}\right)\right]=\dfrac{72}{37}\)

\(\Rightarrow\) \(4x\left[\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)\right]=\dfrac{72}{37}\)

\(4x\left[\dfrac{1}{4}\left(1-\dfrac{1}{37}\right)\right]=\dfrac{72}{37}\)

\(4x\left[\dfrac{1}{4}\left(\dfrac{37}{37}-\dfrac{1}{37}\right)\right]=\dfrac{72}{37}\)

\(4x\left[\dfrac{1}{4}\cdot\dfrac{36}{37}\right]=\dfrac{72}{37}\)

\(4x\cdot\dfrac{9}{37}=\dfrac{72}{37}\)

\(4x=\dfrac{72}{37}:\dfrac{9}{37}\)

\(4x=\dfrac{72}{37}\cdot\dfrac{37}{9}\)

\(4x=8\)

\(x=\dfrac{8}{4}\)

\(x=2\)

Vậy \(x=2\).

\(\dfrac{4x}{5}+\dfrac{4x}{45}+\dfrac{4x}{117}+....+\dfrac{4x}{1221}=\dfrac{72}{37}\)

\(x\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{x}{33.37}\right)=\dfrac{72}{37}\)

\(x\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{33}-\dfrac{1}{37}\right)=\dfrac{72}{37}\)

\(x\left(1-\dfrac{1}{37}\right)=\dfrac{72}{37}\)

\(x\cdot\dfrac{36}{37}=\dfrac{72}{37}\)

\(x=\dfrac{72}{37}:\dfrac{36}{37}\)

\(x=2\)

Số tiếp theo là 151

a. = 100

b. = 7200

c. =10000000

d. = 2500

thanks

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