Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

                  Đúng không bạn

Bạn chỉ nên đưa câu khó lên vì làm cả đề thì ko làm kịp đc

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

a. x - (17 - 8) = 5 + (10 - 3x)

=> x - 9 = 5 + 10 - 3x

=> x + 3x = 15 + 9

=> 4x = 24

=> x = 6

b. 25 - (30 + x) = x - (27 - 8)

=> 25 - 30 - x = x - 19

=> -x -x = -19 + 30 - 25

=> -2x = -14

=> x = 7

c. (x - 12) - 15 = (20 - 7) - (18 + x)

=> x - 12 - 15 = 13 - 18 - x

=> x + x = 13 - 18 + 12 + 15

=> 2x = 22

=> x = 11

Cảm ơn nhé bạn Minh Hiền!

a) 4(x+1)-(3x+1)=14

<=>4x+4-3x-1=14

<=>4x-3x = 14-4

<=>x=10

b) -12(x-5) +7.(3x)=5

<=> -12x +60 + 21+7x=5

<=> -12x+7x=5-60-21

<=> -5x=-76

<=> x=76/5

c) 17x + 34 -4x+28-9x=82

<=> 17x-9x-4x=82-34-28

<=> 4x=20

<=> x=5

cần trình bày cách làm không bạn?

a) x=11

b) x=-55/5

c) x= 5

a) x - (11 - x) = -48 + (-12 + x)

x - 11 + x = -48 + (-12) + x

x + x - x = -48 + (-12) + 11

x + 0 = (-60) + 11

x = -49

b) (15 - x) + (x - 12) = 7 - (-8 + x)

15 - x + x - 12 = 7 + 8 + x

15 - x + x - 12 = 15 + x

x + x - x = 15 - 15 + 12

x + 0 = 0 + 12

x = 12

c) (x - 12) - (2x + 31) = -6 - 5

x - 12 - 2x - 31 = -1

x - 12 - 2x = -1 + 31

x - 12 - 2x = 30

x - 2x = 30 + 12

-1x = 42

=> x = -42

d) |x + 5| - (-17) = 20

=> |x + 5| + 17 = 20

=> |x + 5| = 20 - 17

=> |x + 5| = 3

=> \(\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=3-5\\x=-3-5\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

12x² - 3x = 0

=> 12x² - 3x = (2² . 3x²) - 3x = 0

=> 3x . (4x - 1) = 0

=> 3x . 4x - 3x . 1 = 0

=> 3x . 4x - 3x = 0

=> 12x - 3x = 0

=> 4x  =0

=> x = 0

x=0 chac chan

Thanks bạn nhìu!

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

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

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

a)\(-\frac{12}{7}\cdot\left(\frac{3}{4}-x\right)\cdot\frac{1}{4}=0\)

=>\(\frac{3}{4}-x=0\)

=>\(x=\frac{3}{4}\)

 \(x=\frac{3}{4}\)

b) \(x:\frac{17}{8}=-\frac{2}{5}\cdot\left(-\frac{9}{17}\right)\)

=>\(x:\frac{17}{8}=\frac{18}{85}\)

=>\(x=\frac{18}{85}\cdot\frac{17}{8}=\frac{9}{20}\)

 \(x=\frac{9}{20}\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}