a,=629.0408

b,=3891.3

c,=3725

d,=5104

\(100:\left\{2\cdot\left[22-\left(35-8\right)\right]\right\}\\ =100:\left\{2\cdot\left[22-27\right]\right\}\\ =100:\left(-10\right)\\ =-10\)

\(12:\left\{390:\left[500-\left(125+35\cdot7\right)\right]\right\}\\ =12:\left\{390:\left[500-370\right]\right\}\\ =12:3\\ =4\)

a) 100:{ 2.[22-(35-8)]}

=100:{2.[22-27]}

=100:{2.(-5)}

=100: 10

=10

a) -10

b)4

A = 5,65 \(\times\) 13,8 + 35,7 \(\times\) 3,21 + 5,65 \(\times\) 21,9 + 35,7 \(\times\) 1,14

A = 5,65 \(\times\) (13,8 + 21,9) + 35,7 \(\times\) (3,21 + 1,14)

A = 5,65  \(\times\) 35,7 + 35,7 \(\times\) 4,35

A = (5,65 + 4,35) \(\times\) 35,7

A = 10 \(\times\) 35,7

A = 357

ritug8EB5;7H;98$yb 

A=[5.65 x (13.8+21.9)]+[35.7 x (3.21+1.14)]

  =[5.65 x 35.7]+[35.7 x 4.35]

  =5.65 x 35.7 + 35.7 x 4.35

  =35.7 x (5.65+4.35)

  =35.7 x 10

  =357

 x - 91,5 + 35,7 = 113 , 09 

 x - 91,5 = 113,09 - 35,7 

 x - 91,5 = 77,39

 x = 77,39 + 91,5 

 x = 168,89 

Vậy x = 168,89

x - 91,5 + 35,7 = 113,09

           x - 55,8 = 113,09

                     x = 113,09 + 55,8 

                    x = 168,89

#)Giải :

     x - 91,5 + 35,7 = 113,09

     x - 91,5            = 113,09 - 35,7 

     x - 91,5            = 67,39

     x                     = 67,39 + 91,5

     x                     = 158,89

         #~Will~be~Pens~#

Ta có : \(\frac{x-99}{5}+\frac{x-99}{15}+\frac{x-99}{25}+\frac{x-99}{35}=0\)

\(\Rightarrow\left(x-99\right)\left(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\right)=0\)

Vì \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\ne0\)

Nên : x - 99 = 0

Suy ra : x = 99

Vậy x = 99

Lời giải:

$14.7^{2021}=35.7^{2021}-3.49^x$

$2.7^{2022}=5.7^{2022}-3.7^{2x}$

$3.7^{2x}=5.7^{2022}-2.7^{2022}=7^{2022}(5-2)=3.7^{2022}$

$\Rightarrow 7^{2x}=7^{2022}$

$\Rightarrow 2x=2022$

$\Rightarrow x=2022:2=1011$

\(14.7^{2021}=35.7^{2021}-3.49^x\)

\(\Leftrightarrow2.7^{2022}=5.7^{2022}-3.7^{2x}\)

\(\Leftrightarrow3.7^{2x}=3.7^{2022}\) \(\Leftrightarrow7^{2x}=7^{2022}\)

\(\Leftrightarrow2x=2022\Leftrightarrow x=1011\) (TM \(x\in Z\))

Vậy \(x=1011\)

Sửa đề

\(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)

\(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)

\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)

\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)

\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)

\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)

\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)

\(\dfrac{1}{x+2}=\dfrac{-9}{35}\)

\(x+2=35\)

\(x=35-2\)

\(x=33\)

\(-17x=-17-\left(-34\right)\)

\(-17x=-17\)

\(x=-1\)

Bạn coi lại đề: $\frac{3}{69}$ không nằm trong các số hạng có quy luật.