a) X x 2,8 + X x 5,2 = 48

X x (2,8+5,2)=48

X x 8=48

X=48:8

X=6

b) X x 12,6 -  X x 5,6 =42

X x (12,6-5,6)=42

X x 7=42

X=42:7

X=6

Các câu khác làm tương tự nha

a) x * 2.8 + x * 5.2 = 48 

<=> x * ( 2.8 + 5.2 ) = 48

       x * 8                  = 48

        x                       = 48 : 8

        x = 6

b) x * 12.6 - x * 5.6 = 42

<=> x * ( 12.6 - 5.6 ) = 42

        x *         7          = 42

        x                        = 42 : 7

        x                        = 6

c) x * 12.25 - x + x * 2.75 = 1050

<=> x * ( 12.25 - 1 + 2.75 ) = 1050

        x *        14                   = 1050

        x                                  = 1050 : 14 

        x                                   = 75

d) x * 9.1 + x * 1.9 = 26.4

<=> x * ( 9.1 +1.9 ) = 26.4

        x *       11        = 26.4

        x                      = 26.4 : 11

        x                      = 2.4

\(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}-\frac{1}{110}=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\frac{1}{3}+\frac{1}{11}=x-\frac{5}{13}\)

\(\frac{1}{11}=x-\frac{5}{13}\)

\(x=\frac{1}{11}+\frac{5}{13}\)

\(x=\frac{68}{143}\)

     x - 3/7 - 3/42 - 3/102 - 3/187 = 2

=> x - [ 3× ( 1/7 - 1/42 - 1/102 - 1/187 ) ] = 2

=> x - 24/77 = 2

=> x = 178/77

k mk vs. Chúc bạn học thật giỏi

     x-3/7-3/42-3/102-3/187

=> x=2+3/7+3/42+3/108+3/187

=> x=28/11

Ở trên mk làm lộn . Sorry

k mk vs. Chúc bạn học thật giỏi

mình ko đáng cái j linh tinh hết đây là các bài toán mà mình ko giải đc

b. (x-7)^x+1-(x-7)^x+11=0

(x-7)^x+1.[1-(x-7)¹⁰]=0

=> (x-7)^x+1=0 hoặc 1-(x-7)¹⁰=0

• (x-7)^x+1= 0 => x-7=0 => x=7

• 1-(x-7)¹⁰=0=> (x-7)¹⁰=1=>x-7=1 hoặc x-7=-1 => x=8 hoặc x=6

Vậy x thuộc {6;7;8}

lớp mấy đấy

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

Em cảm ơn ạ.

2:

=>27:3^x=2*25-16-31=50-47=3

=>3^x=27/3=9

=>3^x=3^2

=>x=2

1:

b: \(=16\cdot55+16\cdot45-1\)

=16(55+45)-1

=1600-1

=1599

c: \(=\dfrac{1800}{49-\left[2\cdot\left(36-34\right)^3-5\right]}\)

\(=\dfrac{1800}{49-2\cdot2^3+5}=\dfrac{1800}{49-16+5}=\dfrac{1800}{38}\)=900/19

d: \(=\dfrac{5\cdot3^{11}\cdot2^{11}-3^{11}\cdot2^{11}}{2^{10}\cdot3^{10}\cdot2^2\cdot3+7\cdot2^{12}\cdot3^{12}}\)

\(=\dfrac{3^{11}\cdot2^{11}\left(5-1\right)}{2^{12}\cdot3^{11}\left(1+7\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{4}{1+21}=\dfrac{4}{22\cdot2}=\dfrac{1}{11}\)

\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)

\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))

\(\Leftrightarrow x=-110\)

Vậy \(x=-110\)

mn giải hộ mik ik