=>7x=7

hay x=1

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

=>-11/5x=1/5

=>x=-1/11

lxl=3x+5

lxl=x<=> x ≥ 0

lxl= -x <=> x<0

TH1 lxl=3x+5                                         TH2  lxl=3x+5

x=3x+5 (x ≥ 0)                                        -x=3x+5 (x<0)

-2x=5                                                        -4x =5

x=\(\dfrac{-5}{2}\)KTM                                                 x=\(\dfrac{-5}{4}\)TM

x + 3/4 x 7/10 = 53/40

x + 21/40 = 53/40

x = 53/40 - 21/40

x = 32/40

x = 4/5

34/35 : x = 17/5

x = 34/35 : 17/5

x = 34/35 x 5/17

x = 2/7

x + 3/4 x 7/10 = 53/40

x + ( 3/4 x 7/10 ) = 53/40

x + 21/40 = 53/40 ( tới đây bn cứ lm như bth )

x              = 53/40 - 21/40

x              = 4/5

34/35 : x = 17/5

x             = 34/35 : 17/5

x             = 2/7

\(a,\) \(x+\dfrac{3}{4}\times\dfrac{7}{10}=\dfrac{53}{40}\) 

    \(x+\dfrac{21}{40}=\dfrac{53}{40}\) 

            \(x=\dfrac{53}{40}-\dfrac{21}{40}\) 

            \(x=\dfrac{4}{5}\) 

\(b,\) \(\dfrac{34}{35}:x=\dfrac{17}{5}\) 

            \(x=\dfrac{34}{35}:\dfrac{17}{5}\) 

            \(x=\dfrac{2}{7}\)

`-5/(2x)=-5`

`=> -5=2x.(-5)`

`=> 2x.(-5)=-5`

`=>2x=-5:(-5)`

`=>2x=1`

`=>x=1:2`

`=>x=1/2`

- 5/2x = - 5

- 5/2x = -5x/2

=>-5x/2=-5

=>-5(x-2)/2=0

=>-5(x+2)=0

=>x+2=0

=>x=2

a) 2x² (x+1)+4(x+1)

=(2x²+4)(x+1)

=2(x²+2)(x+1)

b) -3x-6xy+9xz

=9xz-3x-6xy

=3x(3z-2y-1)

c) 2x²y -4xy²+ 6xy

=-2xy(2y-x-3)

Em xem lại đề nhé! Có xuất hiện dấu + không? Hay chỉ là dấu x

A= 4+4\(^2\)+4\(^3\)+4\(^4\)+...+4\(^{2021}\)+4\(^{2022}\)⋮5

A=(4+4\(^2\))+(4\(^3\)+4\(^4\))+...+(4\(^{2021}\)+4\(^{2022}\))⋮5

A=4(1+4)+4\(^2\)(1+4)+...+4\(^{2021}\)(1+4)⋮5

A=4.5+4\(^2\).5+...+4\(^{2021}\).5⋮5

A=(4+4\(^2\)+...+4\(^{2021}\)).5⋮5

Vậy A⋮5

\(A=4+4^2+4^3+4^4+...+4^{2021}+4^{2022}\)

\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)

\(=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{2021}.\left(1+4\right)\)

\(=4.5+4^3.5+...+4^{2021}.5\)

\(=5.\left(4+4^3+...+4^{2021}\right)⋮5\)

Vậy \(A⋮5\)

2x^3+6x^2=x^2+3x

=>2x^3+6x^2-x^2-3x=0

=>2x^3+5x^2-3x=0

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

=>x=0 hoặc 2x^2+5x-3=0

=>x=0  ,       2x^2-x+6x-3=0

=>x=0  ,        x(2x-1)+3(2x-1)=0

=>x=0  ,        (2x-1)(x+3)=0

=>x=0  ,   2x-1=0 hoac x+3=0

=> x=0 ,   x=1/2   ,x=-3

giải ra giùm cái đy chi tiết vô