a, X = 273

b, X = 156

c, X = 4

d, X =  505

e, X = \(\frac{37}{245}\)

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

d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\) 

l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)

Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)

c) x = 3,5 hoặc -3,5

d) x = -2,7

e) x = -1,27

f) x = 0

g) x = -2

h) x = 0

l) x = 9/4

   x = -15/4

Cần bổ sung điều kiện \(x;y\inℤ\)

a) \(\left(x-7\right)\left(xy+1\right)=9=1\cdot9=3\cdot3=\left(-1\right)\cdot\left(-9\right)=\left(-3\right)\cdot\left(-3\right)\)

Xét bảng :

Vì x,y thuộc Z nên ta có (x;y)={(8;1),(16;0),(-2;1),(4;-1)

b) \(\left(x+5\right)\left(3x-12\right)>0\)

TH1 : \(\hept{\begin{cases}x+5>0\\3x-12>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-5\\x>4\end{cases}\Leftrightarrow x>4}}\)

TH2 : \(\hept{\begin{cases}x+5< 0\\3x-12< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -5\\x< 4\end{cases}\Leftrightarrow x< -5}}\)

Vậy....

c) \(\left|x+3\right|< 3\)

Mặt khác : \(0< \left|x+3\right|< 3\)

\(\Rightarrow\left|x+3\right|\in\left\{1;2\right\}\)

\(\Rightarrow\left(x+3\right)\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)

Vậy....

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

a) \(19\times\left(52+38\right)-9\times\left(52+38\right)=\)

\(=\)     \(19\times90-9\times90\) 

\(=90\times\left(19-9\right)\) 

\(=90\times10\) 

\(=900\) 

b) \(\dfrac{2}{5}\div\dfrac{3}{7}\times\dfrac{3}{7}\div\dfrac{2}{5}+2019=\)

\(=\left(\dfrac{2}{5}\div\dfrac{2}{5}\right)\times\left(\dfrac{3}{7}\div\dfrac{3}{7}\right)+2019\) 

\(=1\times1+2019\) 

\(=2019\)

Bài 5 :

S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399

S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)

S = 1 + 0 + 0 + ... + 0 + (- 401)

S = 1 - 401

S = - 400

Bài 5

A= 1+3-5-7+9+11-13-15+...-397-399

A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)

A= -8 -8 -...-8

A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)

A= -400

Bài 6: 

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: S={0;-3}

c) Ta có: 52-(x+21)=19-x

\(\Leftrightarrow52-x-21-19+x=0\)

\(\Leftrightarrow12=0\)(vô lý)

Vậy: \(x\in\varnothing\)

a)x-63:9=18

   x-63  =18.9

   x-63  =162

   x      =162+63

   x      =225

b)3(29-x)+52=103

   3(29-x)     =103-52

   3(29-x)     =51

      29-x      =51:3     

      29-x      =17

          x       =12

c)|x-1|=4

=>x-1=4 hoặc x-1= -4

=>x=5 hoặc x= -3

d)(-7)-x= -21

        x=(-7)-(-21)

        x=14

a) x-63;9=18

 x-7=18

x=18+7=25

b)3(29-x)=103-52

87-3x=51

    -3x=51-87=-36

     x=12

c)

* x-1 =4

x=5

* x-1=-4

x=-3

d)(-7)-x=-21

-x=-21+7=14

x=14

a x - 4/5 = 7/25

x = 7/25 + 4/5

x = 27/25

b x :3/5 = 7/9

x = 7/9 x 3/5

x = 7/15

c x x 15 = 2850

x = 2850 : 15

x = 190

d x : 52 = 113

x = 113 x 52

x = 5876

a: x=7/25+4/5=7/25+20/25=27/25

b: x:3/5=7/9

nên x=7/9x3/5=21/45=7/16

c: 15x=2850

nên x=190

`a)x=7/25+4/5`

`=7/25+20/25`

`=27/25`

`b)x:3/5=7/9`

`x=7/9x3/5`

`x=7/16`

`c)x xx 15 = 2850`

`x = 2850 : 15`

`x = 190`

`d)x = 113 xx 52`

`x = 5876`