A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

A, \(\dfrac{x}{5}=\dfrac{2}{5}\Rightarrow x=2\)

B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)

C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)

D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)

   \(\Rightarrow2x=6;6=x:2;x=3\)

a)12 x 18 + 14 x 3 - 255 : 17=25.58823529411765

b)68 + 42 x 5 - 625 : 25=5182

c) 13+ 21 x 5 - (198 : 11 - 8)=108

d) 1188 - 4 x (27 + 90 + 73 :10 )=

e) 5 + 8 + 11 + 14 +......+ 38 + 41=

f) 51 - 49 + 55 - 53 + 59 - 56 + 63 - 61 + 65=74

có mấy câu mình ko biết,xin lỗi bạn

a) \(3\left(2x-5\right)+125=134\)

\(\Leftrightarrow3\left(2x-5\right)=9\)

\(\Leftrightarrow2x-5=3\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)

\(\Leftrightarrow6x+9=27\)

\(\Leftrightarrow6x=18\Leftrightarrow x=3\)

d) \(27\left(x-27\right)-27=0\)

\(\Leftrightarrow27\left(x-27\right)=27\)

\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)

3.(x+5)-2x=27
3.x+3.5-2x=27
3x+15-2x=27
x.(3-2)+15=27
x.  1    +15=27
  x       +15=27
  x             =27-15
  x             =12

= 4/3 x 6/5 + 3/10

= 8/5 +3/10

= 19/10

cám ơn nha

k mik nha

a) Ta có: \(\left(x-3\right)\left(1-x\right)-2\)

\(=-x^2+4x-3-2\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1< 0\left(\forall x\right)\)

=> đpcm

b) Ta có: \(\left(x+4\right)\left(2-x\right)-10\)

\(=-x^2-2x+8-10\)

\(=-\left(x^2+2x+1\right)-1\)

\(=-\left(x+1\right)^2-1\le-1< 0\left(\forall x\right)\)

=> đpcm

b) \(\left(x+4\right)\left(2-x\right)-10=-x^2-4x+2x+8-10\)

\(=-x^2-2x-2=-x^2-2x-1-1\)

\(=-\left(x^2+2x+1\right)-1=-\left(x+1\right)^2-1\)

Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+2\right)^2-1\le-1\forall x\)

\(\Rightarrow\left(x+4\right)\left(2-x\right)-10\le-1\forall x\)

hay \(\left(x+4\right)\left(2-x\right)-10\)luôn âm với mọi x ( đpcm )

a) (x-3).(1-x)-2

= x - x²-3+3x-2

=4x-x²-5

= - ( x²-4x+5)

= -(x²-2.2x+4)-1

= - (x-2)²-1 

Có : - ( x-2)² \(\le\)0 với mọi x

=> -(x-2)²-1\(\le\)-1 < 0 với mọi x

Vậy ...

b) (x+4).(2-x)-10

2x-x²+8-4x-10

= -2x -x² - 2

= -(x²+2x+2)

=-(x²+2x+1)-1

=-(x+1)²-1 

rồi đến đây lập luận như câu a nhé . Tớ lười lắm T_T thông cảm

a. $\dfrac{14}{15}+\dfrac{2}{11}+\dfrac{1}{15}+\dfrac{4}{11}+\dfrac{10}{22}$

$=(\dfrac{14}{15}+dfrac{1}{15})+(\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{10}{22})$

$=1+1=2$

b. $74\times43-43+27\times43$

$=(74-1+27)\times43$

$=100\times43=4300$

\(a)\dfrac{14}{15}+\dfrac{2}{11}+\dfrac{1}{15}+\dfrac{4}{11}+\dfrac{10}{22}\)

\(=\left(\dfrac{14}{15}+\dfrac{1}{15}\right)+\dfrac{2}{11}+\dfrac{4}{11}+\dfrac{5}{11}\)

\(=1+1=2\)

\(b)74\times43-43+27\times43\)

\(=43\times\left(74-1+27\right)\)

\(=43\times100=4300\)

a) \(\dfrac{14}{15}+\dfrac{2}{11}+\dfrac{1}{15}+\dfrac{4}{11}+\dfrac{10}{22}\)

= \(\left(\dfrac{14}{15}+\dfrac{1}{15}\right)+\left(\dfrac{2}{11}+\dfrac{4}{11}\right)+\dfrac{5}{11}\)

= \(1+\dfrac{6}{11}+\dfrac{5}{11}\)

= 2

b) 74 x 43 - 43 + 27 x 43

= 74 x 43 - 43 x 1 + 27 x 43

= 43 x ( 74 - 1 + 27 )

= 43 x 100

= 4300

a) x=-213:(1+2+3+4+...+100)<=>x=-213/100

b) x-x=-1/3-2/4 <=> 0= -5/6 (vô lý )

c) x=-0,8119408369

d) x= 0.0258907758

help me 

x+(x+1)+(x+2)+...+1008=2008

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)

    \(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)

     \(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)