câu a : 

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\\ \left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=\dfrac{7}{4}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}+\dfrac{1}{2}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\\ x-\dfrac{1}{3}=\sqrt{\dfrac{9}{4}}\\ x-\dfrac{1}{3}=\dfrac{3}{2}\\ x=\dfrac{3}{2}+\dfrac{1}{3}\\ x=\dfrac{11}{6}\)

câu b : 

\(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\\ \Rightarrow\left(x-3\right)\cdot\left(x-3\right)=\left(-2\right)\cdot\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ x-3=\sqrt{16}\\ x-3=4\\ x=4+3\\ x=7\)

câu c : 

\(\dfrac{9}{x}=\dfrac{-35}{105}\\ \Rightarrow\left(-35\right)\cdot x=9\cdot105\\ \left(-35\right)\cdot x=945\\ x=945\div\left(-35\right)\\ x=-27\)

  ( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

  ( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

   (\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

    (\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\)  = \(\dfrac{1}{18}\)

     ( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

       (\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

         \(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)

               \(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)

               \(x\) = 4

x = 4 nhé

 ( 215152​ + 235352​ + 263632​) : �x = 118181​

  ( 2×715×715×72×7​ + 2×335×335×32×3​ + 263632​) : �x = 118181​

   (1410510514​ + 61051056​ + 263632​) : �x = 118181​

    (2010510520​ + 263632​) : �x  = 118181​

     ( 421214​ + 263632​) : �x = 118181​

       (12636312​ + 263632​) : �x = 118181​

         2992​ : �x = 118181​

               �x = 2992​ : 118181​

               �x = 4

a,     \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\)    = \(\dfrac{10}{21}\)

    (\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\)  =  \(\dfrac{10}{21}\)

     - \(\dfrac{5}{21}\) \(\times\) \(x\)      = \(\dfrac{10}{21}\)

                 \(x\)      = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))

                 \(x\)      = -2 

b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)

            \(x\) - \(\dfrac{1}{3}\)    =  \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))

             \(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)

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

              \(x\)      = - \(\dfrac{13}{6}\)

c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)

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

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

              \(x\)           = - \(\dfrac{47}{10}\)

\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)

\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)

a,73​x−32​x=2110​⇒x(73​−32​)=2110​⇒x.−215​=2110​⇒x=−2b,357​:(x−31​)=−252​⇒51​:(x−31​)=−252​⇒x−31​=−25​⇒x=−613​c,3.(x−21​)−5.(x+53​)=−x+51​⇒3x−23​−5x+5=−x+51​

⇒�(3−5)−32+5=−�+15⇒−2�−132=−�+15⇒−�−135=15⇒�=15−135⇒�=−125.⇒x(3−5)−23​+5=−x+51​⇒−2x−213​=−x+51​⇒−x−513​=51​⇒x=51​−513​⇒x=−512​.

Lời giải:

a. $\frac{2-x}{4}=\frac{3x-1}{3}$

$\Rightarrow 3(2-x)=4(3x-1)$

$\Rightarrow 6-3x=12x-4$

$\Rightarrow 6+4=12x+3x$

$\Rightarrow 10=15x$

$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$

b.

$\frac{x}{7}=\frac{x+16}{35}$

$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$

$\Rightarrow 5x=x+16$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

c.

$\sqrt{x^2+1}=3$

$\Rightarrow x^2+1=9$

$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$

2: 

a: =>x=1/2+1/6=3/6+1/6=4/6=2/3

b: =>x+3,5=7,8-6,3=1,5

=>x=1,5-3,5=-2

c: =>(35-x)/-105=1/5

=>35-x=-21

=>x=56

1

c hình như tính bình thường thôi:v

d

= 1,75 : 5 + 2,5 . (16 – 4 . 4,1)

= 1,75 : 5 + 2,5 . (16 – 16,4)

= 1,75 : 5 + 2,5 . (−0,4)

= 0,35 − 1

= −0,65.

2

a

\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

b

\(\Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{5}:2\\ \Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{10}=\dfrac{15}{10}\\ \Rightarrow x=\dfrac{15}{10}-\dfrac{35}{10}=-2\)

c không rõ đề:v

\(1,\dfrac{2}{-11}.6\dfrac{2}{7}+\dfrac{3}{7}:3+5\\ =\dfrac{2}{-11}.\dfrac{44}{7}+\dfrac{3}{7}\times\dfrac{1}{3}+5\\ =-\dfrac{88}{77}+\dfrac{3}{21}+5\\ =-\dfrac{8}{7}+\dfrac{1}{7}+5\\ =-1+5\\ =4\)

\(d,1,75:5+2,5.\left(4^2-4.4,1\right)\\ =0,35+2,5.\left(16-16,4\right)\\ =0,35+2,5.0,4\\ =0,35+1\\ =1,35\)

`2,`

`a, 1/2 -x=-1/6`

`=> x= 1/2 -(-1/6)`

`=> x= 1/2 +1/6`

`=> x= 3/6+1/6`

`=>x= 4/6`

`=>x=2/3`

Vậy `x=2/3`

`b, x+3,5 =7,8 -12,6 :2`

`=> x+3,5 = 7,8 - 6,3`

`=> x+3,5=1,5`

`=> x= 1,5-3,5`

`=>x=-2`

Vậy `x=-2`

\(c,\dfrac{7}{35}=\dfrac{35-x}{-105}\\ \Rightarrow7.\left(-105\right)=35.\left(35-x\right)\\ \Rightarrow-735=1225-35x\\ \Rightarrow-735-1225=-35x\\ \Rightarrow-1960=-35x\\ \Rightarrow x=-1960:\left(-35\right)\\ \Rightarrow x=56\)

Vậy `x=56`

e: =>3x-6=12x-4

=>-9x=2

=>x=-2/9

f: =>5x=x+16

=>4x=16

=>x=4

b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)

\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)

\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)

Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)

\(\Rightarrow t^2-t-20=0\)

\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)

Thay t=5 vào (1), ta có :

\(\sqrt{x^2-3x+7}=5\)

\(\Leftrightarrow x^2-3x+7=25\)

\(\Leftrightarrow x^2-3x-18=0\)

\(\Rightarrow x_1=6;x_2=-3\)

vậy...

xl bn tớ gửi nhầm

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

=>\(\dfrac{x+9-x-1}{\left(x+9\right)\left(x+1\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow2\left(x^2+10x+9\right)=5\cdot8=40\)

=>x^2+10x+9=20

=>x^2+10x-11=0

=>(x+10)(x-1)=0

=>x=1 hoặc x=-10

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{2}{\left(x+3\right)\left(x+5\right)}+\dfrac{2}{\left(x+5\right)\left(x+7\right)}+\dfrac{2}{\left(x+7\right)\left(x+9\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{x+9-x-1}{\left(x+1\right)\left(x+9\right)}=\dfrac{2}{5}\)

=>2(x+1)(x+9)=5*8=40

=>x^2+9x+9=20

=>x^2+9x-11=0

hay \(x=\dfrac{-9\pm5\sqrt{5}}{2}\)

=>x^2+9x