1. 234 + ( 345 - x ) = 500

345 - x = 500 - 234

345 - x = 266

x = 345 - 266

x = 79

2. 456 - ( x + 23 ) = 326

x + 23 = 456 - 326

x + 23 = 130

x = 130 - 23

x = 107

3. ( 567 - x ) . 212 = 605

567 - x = 605 : 212

567 - x = ...

=> Đề sai

4. ( 5 . x - 15 ) : 5 = 0

5 . x - 15 = 0 : 5

5 . x - 15 = 0

5 . x = 0 + 15

5 . x = 15

x = 15 : 5

x = 3

5. 14 x ( x - 7 ) = 210

x - 7 = 210 : 14

x - 7 = 15

x = 15 + 7

x = 22

6. 84 - 4 x ( 2 . x + 1 ) = 48

4 x ( 2 . x + 1 ) = 84 - 48

4 x ( 2 . x + 1 ) = 36

2 . x + 1 = 36 : 4

2 . x + 1 = 9

2 . x = 9 - 1

2 . x = 8

x = 8 : 2

x = 4

1. 345 - x = 500 - 234

345-x=266

x=345-266

x=79

2. x+23 =456-326

x+23=130

x=130-23

x=107

4. 5 x X - 15 =5 x0 

5xX -15=0

5 x X = 0+15

5x X = 15

X = 15:5

X =3

5 . x-7 = 210 :14

x - 7 =15

x=15+7

x=22

\(a,\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(c,\left(x+3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)

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

\(e,\left(x-4\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

\(f,\left(2x-1\right)\left(3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

`a,(x-1)(x+2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

`b,(x -2)(x -5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)

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

`c,(x +3)(x -5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)

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

`d,(x + 1/2)(4x + 4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)

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

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

`e,(x -4)(5x -10)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

`f,(2x -1)(3x +6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)

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

`g,(2,3x -6,9)(0,1x -2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)

a.(x - 1)(x + 2)= 0

<=> x-1=0 hoặc x+2=0

<=> x=1 hoặc x=-2

b.(x -2)(x -5)=0

<=> x-2=0 hoặc x-5=0

<=> x=2 hoặc x=5

c.(x +3)(x -5)=0

<=> x+3=0 hoặc x-5=0

<=> x=-3 hoặc x=5

d.(x + 1/2)(4x + 4)=0

<=> x+1/2=0 hoặc 4x+4=0

<=> x=-1/2 hoặc x=-1

e.(x -4)(5x -10)=0

<=> x-4=0 hoặc 5x-10=0

<=> x=4 hoặc x=2

f.(2x -1)(3x +6)=0

<=> 2x-1=0 hoặc 3x+6=0

<=> x=1/2 hoặc x=-2

g.(2,3x -6,9)(0,1x -2)=0

<=> 2,3x-6,9=0 hoặc 0,1x-2=0

<=> x=3 hoặc x=20

Bài 1 :

a) ( 257 x 139 - 257 x 39 ) : 100

=  257 x ( 139 - 39 ) : 100

= 257 x 100 : 100

= 257

a) ( 257 × 139 – 257 × 39 ) : 100

= [ 257 x ( 139 -39 ) ] : 100

= ( 257 x 100 ) :100

= 25700 :100

=257

b) 5 + 6 + 13 + 17 + .......+ 2017

= ( 5 + 2017 ) + ( 6 + 2016 ) + ( 7 + 2015 ) + ... + ( 1009 + 1013 ) + ( 1010 + 1012 ) + 1011

= 2022 + 2022 + 2022 + ... + 2022 + 2022 + 1011

= 2022 × 1006 + 1011

= 203514

c) 12 × 57 + 57 × 15 + 63 × 57 

= 57 × ( 12 + 15 + 63 )

= 57 x 90

= 5130

Bài 2:

a) 541 + ( 218 - x ) = 735                          b)  ( x - 50 ) : 25 = 8                                      c)  ( x - 32 ) - 15 = 0

              ( 218 - x ) = 735 - 541                       ( x - 50 )       = 8 x 25                                    ( x - 32 )        = 0 + 15

              ( 218 - x ) = 194                                  x - 50         = 200                                           x - 32          = 15

                          x   = 218 - 194                        x                = 200 + 50                                    x                 = 15 + 32

                          x   = 24                                   x                = 250                                            x                 = 47

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

=>(x-2)(3x-2)=0

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

=>x(x-3)(x+1)=0

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)

e: =>x(x^3-4x^2-8x+8)=0

=>x[(x^3+8)-4x(x+2)]=0

=>x(x+2)(x^2-2x+4-4x)=0

=>x(x+2)(x^2-6x+4)=0

=>\(x\in\left\{0;-2;3+\sqrt{5};3-\sqrt{5}\right\}\)

g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0

=>(2x+5)(x^3-3x^2+3x-1)=0

=>(2x+5)(x-1)^3=0

=>x=1 hoặc x=-5/2

h: =>(x^2+8x+7)(x^2+8x+15)+15=0

=>(x^2+8x)^2+22(x^2+8x)+120=0

=>(x^2+8x+10)(x^2+8x+12)=0

=>(x^2+8x+10)(x+2)(x+6)=0

=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)

a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)

=> x=-1  

với \(3x^2+x-2=0\)

ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)

Vậy  ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

hay \(x\in\left\{1;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)

hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)

sao câu 2+2:2 không có dấu = vậy

có vài câu không phải toán lớp 9 đâu

tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, 3x - 7 = 0

<=> 3x = 7

<=> x = 7/3

b, 8 - 5x = 0

<=> -5x = -8

<=> x = 8/5

c, 3x - 2 = 5x + 8

<=> -2x = 10

<=> x = -5

e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)

`a ) 3x - 7 = 0`

`\(\Leftrightarrow \) 3x = 7`

`\(\Leftrightarrow \) x = 7/3`

Vậy `S = {-7/3}`