\(a,\) \(5.\left(4+6x\right)=290\)

\(4+6x=290:5\)

\(4+6x=58\)

\(6x=58-4\)

\(6x=54\)

\(x=54:6\)

\(x=9\left(tm\right)\)

Vậy ..................

\(b,x.3,7+x.6,3=120\)

\(x\left(3,7+6,3\right)=120\)

\(10x=120\)

\(x=120:10\)

\(x=12\left(tm\right)\)

Vậy ....................

c) \(\left(15.24-x\right):0,25=100:\dfrac{1}{4}\)

\(\left(360-x\right):\dfrac{1}{4}=400\)

\(360-x=400.\dfrac{1}{4}\)

\(360-x=100\)

\(x=360-100\)

\(x=260\left(tm\right)\)

Vậy ...................

a) 5. ( 4+6 . x ) = 290

( 4+6 . x ) = 290 : 5

4 + 6 . x = 58

6 .x = 58 - 4

6.x = 54

x = 54 : 6

x = 9

b) x . 3,7 + x . 6,3 = 120

x ( 3,7 + 6,3 ) = 120

x . 10 = 120

x = 120 : 10

x = 12

 X x ( 3,7 + 6,3 ) = 120

X x 10 = 120

X = 120 : 10

X = 12

Vậy X = 12

a 48

b 

c 

tự làm đừng hỏi

4 + 6 x X = 290 : 5

4 + 6 x X = 58

6 x X = 58 - 4 

6 x X = 54 

X = 54 : 6

X = 9

Vậy X = 9

Giải:

\(9-3\times\left(x-9\right)=6\) 

      \(3\times\left(x-9\right)=9-6\) 

      \(3\times\left(x-9\right)=3\) 

               \(x-9=3:3\) 

               \(x-9=1\) 

                     \(x=1+9\) 

                     \(x=10\) 

\(4+6\times\left(x+1\right)=70\) 

      \(6\times\left(x+1\right)=70-4\) 

      \(6\times\left(x+1\right)=66\) 

               \(x+1=66:6\) 

               \(x+1=11\) 

                     \(x=11-1\) 

                     \(x=10\) 

\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\) 

         \(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\) 

         \(\dfrac{x}{13}=\dfrac{4}{13}\) 

\(\Rightarrow x=4\) 

\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\) 

         \(\dfrac{3}{x}=\dfrac{3}{7}\) 

\(\Rightarrow x=7\) 

\(5\times\left(3+7\times x\right)=40\) 

         \(3+7\times x=40:5\) 

         \(3+7\times x=8\) 

                \(7\times x=8-3\) 

                \(7\times x=5\) 

                      \(x=5:7\) 

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

\(x\times6+12:3=120\) 

       \(x\times6+4=120\) 

             \(x\times6=120-4\) 

             \(x\times6=116\) 

                   \(x=116:6\) 

                   \(x=\dfrac{58}{3}\) 

\(x\times3,7+x\times6,3=120\) 

    \(x\times\left(3,7+6,3\right)=120\) 

                  \(x\times10=120\) 

                           \(x=120:10\) 

                           \(x=12\) 

\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\) 

      \(\left(360-x\right):0,25=400\) 

                   \(360-x=400.0,25\) 

                   \(360-x=100\) 

                             \(x=360-100\) 

                             \(x=260\) 

\(71+65\times4=\dfrac{x+140}{x}+260\) 

\(\left(x+140\right):x+260=71+260\) 

\(x:x+140:x+260=331\) 

    \(1+140:x+260=331\) 

                    \(140:x=331-1-260\) 

                    \(140:x=70\) 

                             \(x=140:70\) 

                             \(x=2\) 

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\) 

                      \(10\times x+\left(1+4+7+...+28\right)=155\)

Số số hạng \(\left(1+4+7+...+28\right)\) :

         \(\left(28-1\right):3+1=10\) 

Tổng dãy \(\left(1+4+7+...+28\right)\) :

         \(\left(1+28\right).10:2=145\) 

\(\Rightarrow10\times x+145=155\) 

               \(10\times x=155-145\) 

               \(10\times x=10\) 

                       \(x=10:10\) 

                       \(x=1\) 

Đều theo cách lớp 5 nha em!

\(\left(84,6-2\cdot x\right):3,02=5,1\)

\(\Rightarrow84,6-2\cdot x=15,402\)

\(\Rightarrow2\cdot x=69,198\)
\(\Rightarrow x=69,198:2\)

\(\Rightarrow x=34,599\)

_____________

\(\left(15\cdot24-x\right):0,25=100:0,25\)

\(\Rightarrow\left(360-x\right):0,25=400\)

\(\Rightarrow360-x=100\)

\(\Rightarrow x-360-100\)

\(\Rightarrow x=260\)

______________

\(128\cdot x-12\cdot x-16\cdot x=5200\)

\(\Rightarrow x\cdot\left(128-12-16\right)=5200\)

\(\Rightarrow x\cdot100=5200\)

\(\Rightarrow x=5200:100\)

\(\Rightarrow x=52\)

__________________

\(5\cdot x+3,75\cdot x+1,25\cdot x=20\)

\(\Rightarrow x\cdot\left(5+3,75+1,25\right)=20\)

\(\Rightarrow10\cdot x=20\)

\(\Rightarrow x=20:10\)

\(\Rightarrow x=2\)

\(x\cdot3,7+x\cdot6,3=360:120\)

\(\Rightarrow x\cdot\left(3,7+6,3\right)=3\)

\(\Rightarrow x\cdot10=3\)

\(\Rightarrow x=\dfrac{3}{10}\)

__________________

\(x\cdot23-6\cdot23+x\cdot69=320\)

\(\Rightarrow x\cdot\left(23+69\right)=320+6\cdot23\)

\(\Rightarrow x\cdot92=458\)

\(\Rightarrow x=458:92\)

\(\Rightarrow x=\dfrac{229}{46}\)

___________________

\(\left(x+1\right)\left(x+2\right)=72\)

\(\Rightarrow x^2+2x+x+2=72\)

\(\Rightarrow x^2+3x+2=72\)

\(\Rightarrow x^2+3x+2-72=0\)

\(\Rightarrow x^2+3x-70=0\)

\(\Rightarrow x^2+10x-7x-70=0\)

\(\Rightarrow\left(x-7\right)\left(x+10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-10\end{matrix}\right.\)

___________________

\(\left(x+2\right)\cdot16\cdot x=160x\)

\(\Rightarrow16x^2+32x=160x\)

\(\Rightarrow16x^2+32x-160x=0\)

\(\Rightarrow16x^2-128x=0\)

\(\Rightarrow16x\left(x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

Đổi:

f.x.23-6.23+x.69=230

=>23.(x-6+3x)=230

=>4x-6=10

=>4x=10+6

=>4x=16

=>x=4

h.(x+2).16.x=160.x

=>(16x+32).x = 160x

=>16x^2 + 32x = 160x

=>16x^2 = 128x

=>16x^2 - 128x=0

=>x(16x-128)=0

=>x=0 hoặc 16x-128=0

+)x=0=>x=0

+)16x-128=0

=>16x=128

=>x=8

     Bài 1:

\(x\) \(\times\) \(\dfrac{15}{16}\) - \(x\) \(\times\) \(\dfrac{4}{16}\) = 2

\(x\) \(\times\) (\(\dfrac{15}{16}\) - \(\dfrac{4}{16}\)) = 2

\(x\) \(\times\) \(\dfrac{11}{16}\) = 2

\(x\) = 2 : \(\dfrac{11}{16}\)

\(x\) = 2 x \(\dfrac{16}{11}\)

\(x\) = \(\dfrac{32}{11}\)

Bài 2: 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{x\left(x+1\right):2}\) = 1 : \(\dfrac{2011}{2012}\)

1 + 2\(\times\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times6}\) + \(\dfrac{1}{2\times10}\) + ... + \(\dfrac{2}{2\times x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\)(\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\) (\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + ... + \(\dfrac{1}{x\times\left(x+1\right)}\)) = 1 + \(\dfrac{1}{2011}\)

1 + 2\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

 1 + 2 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

  1 + 1 - \(\dfrac{2}{x+1}\) = 1 + \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) = 1 + 1 - 1  - \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) =  2 - 1 - \(\dfrac{1}{2011}\)

  \(\dfrac{2}{x+1}\)  = 1 - \(\dfrac{1}{2011}\)

     \(\dfrac{2}{x+1}\) = \(\dfrac{2010}{2011}\)

       \(x\) + 1 = 2 : \(\dfrac{2010}{2011}\)

        \(x\) + 1 = \(\dfrac{2011}{1005}\)

         \(x\) = \(\dfrac{2011}{1005}\) - 1  = \(\dfrac{1006}{1005}\)(loại vì \(\dfrac{1006}{1005}\) không phải là số tự nhiên)

Vậy không có giá trị nào của \(x\) là số tự nhiên thỏa mãn đề bài. 

Bài 3:

\(\dfrac{x}{16}\) \(\times\) (2017 - 1) = 2

\(\dfrac{x}{16}\) \(\times\) 2016 = 2

 \(\dfrac{x}{16}\)             = 2 : 2016

 \(\dfrac{x}{16}\)            = \(\dfrac{1}{1008}\)

   \(x\)             = \(\dfrac{1}{1008}\) x 16

  \(x\) = \(\dfrac{1}{63}\)

\(x+xy+x=4\)

\(\Leftrightarrow x\left(2+y\right)=4\)

Mà \(x,y\inℤ\Rightarrow2+y\inℤ\)

Do đó, \(x,2+y\) là các cặp ước của 4.

Ta có bảng sau :

Vậy : \(\left(x,y\right)\in\left\{\left(-1,-6\right);\left(1,2\right);\left(2,0\right);\left(-2,-4\right);\left(4,-1\right);\left(-4,-3\right)\right\}\)

\(\Leftrightarrow\)x(1+y+1)=4

\(\Leftrightarrow\)x(2y)=4

\(\Rightarrow\)x(2y)\(\in\)Ư4 =1,4,2,-1,-2,-4

lâp bảng

x=1\(\Rightarrow\)y=2

x=2\(\Rightarrow\)y=1

x=4\(\Rightarrow\)y= không có giá trị nào

x=-1\(\Rightarrow\)y=-2

x=-2\(\Rightarrow\)y=-1

x=-4\(\Rightarrow\)y= không có giá trị nào

\(x+xy+x=4\)

\(\Leftrightarrow2x+xy=4\)

\(\Leftrightarrow x.\left(y+2\right)=4\)

Ta lập bảng :

\(Vậy:............\)

5 x X + 3,75 x X + 1,25 x X = 20     

X x (5+3,75+1,25) =20

X x 10 =20

X=20:10

X=2

5 x \(x\)+3,75 x \(x\)+1,25 x \(x\)=20

=> \(x\)x(5+3,75+1,25)=20

=>\(x\)x 10=20

=>x=20:10

=> x=2

vậy \(x\)=2

5 x X + 3,75 x X + 1,25 x X = 20

(5 + 3,75 + 1,25) x X = 20

10 x X = 20

X = 20 : 10

X = 2

*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)

*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)