\(\left|2x^2-27\right|^{2019}+\left(5y+12\right)^{2018}=0.\)

\(\text{Ta có}\hept{\begin{cases}\left|2x^2-27\right|^{2019}\ge0\\\left(5y+12\right)^{2018}\ge0\end{cases}}\text{Mà}\left|2x^2-27\right|^{2019}+\left(5y+12\right)^{2018}=0\)

\(\Rightarrow\hept{\begin{cases}\left|2x^2-27\right|^{2019}=0\\\left(5y+12\right)^{2018}=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(2x-27\right)^{2019}=0\\\left(5y+12\right)^{2018}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-27=0\\5y+12=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=27\\5y=-12\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{27}{2}\\y=\frac{-12}{5}\end{cases}}}}}}\) 

\(\text{Vậy}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-12}{5}\end{cases}}\) 

2x-2016=2018

2x        =2018+2016

2x        =4034

x          =4034:2

x          =2017

3^x +1=82

3^x     =82-1

3^x     =81

3^x     =3⁴

x      =4

20x-11=20x+2x-12

20x-11=x(20+2)-12

20x-11=22x-12

20x    =22x-12+11

20x    =22x-1

22x-2x=22x-1

2x       =1

x         =1:2

x         =0,5

(x+1)+(x+2)+(x+3)+(x+4)=26

x+x+x+x+1+2+3+4        =26

4x+10                           =26

4x                                =26-10

4x                                =16

x                                  =16:4

x                                  =4

I 2x-3 I = I x+1 I

2x-3 = x+1

x+1 - 2x+3=0

x (1-2) +1+3=0

-1x +4 =0

-1x      = 0-4

-1x      =-4

x          = -4 : -1

x         =4

Trả lời:

    \(\left|2x-3\right|=\left|x+1\right|\)

\(\Rightarrow2x-3=x+1\) hoặc   \(2x-3=-\left(x+1\right)\)

TH1:   \(2x-3=x+1\)

           \(2x-x=1+3\)

            \(x=4\)

TH2: \(2x-3=-\left(x+1\right)\)

         \(2x-3=-x-1\)

          \(2x+x=-1+3\)

          \(3x=2\)

          \(x=\frac{2}{3}\)

          Vậy \(x=4;x=\frac{2}{3}\)

\(\left|\frac{1}{3}-x\right|-\left|-\frac{2}{3}\right|=\left|\frac{3}{4}\right|\)

\(\Leftrightarrow\frac{1}{3}-x-\frac{2}{3}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{3}-x=\frac{3}{4}+\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{3}-x=\frac{17}{12}\)

\(\Leftrightarrow x=\frac{1}{3}-\frac{17}{12}=\frac{4}{12}-\frac{17}{12}=\frac{-13}{12}\)

\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)

\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)

Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên 2x + 4 = 0

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

Vậy, x = -2

\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)

\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)

Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên \(2x+4=0\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

Vậy, x = -2

a, 2x+1 chia hết cho x-1

=>2x-2+3 chia hết cho x-1

=>2(x-1)+3 chia hết cho x-1

=>3 chia hết cho x-1

=>x-1 E Ư(3)={1;-1;3;-3}

=>x E {2;0;4;-2}

b, 3x+2 chia hết cho 2x-1

=>2(3x+2)-3(2x-1) chia hết cho 2x-1

=>6x+4-6x-3 chia hết cho 2x-1

=>1 chia hết cho 2x-1

=>2x-1 E Ư(1)={1;-1}

=>x E {1;0}

cảm ơn bạn nhìu nha

a,x²-25-(x+5)                                                   b,mình quên mất rồi.Đợi tí nhé

(x²-25)-(x+5)=0

(x²-5²)+(x-5)=0

(x-5)(x+5)+(x-5)=0

(x-5)(x+5+1)=0

x-5=0 hoặc x+5+1=0

x=0+5 hoặc x=0-5-1

x=5     hoặc x=-6

Vậy x=5 và x=-6

Giải bpt

A)  (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)

B) (x-2)×x^2>0

Mog mn giúp ạ

E cần gấp

Thak mn