1.

a, => 21-x+3 < 0

=> 24-x < 0

=> x < 24

b, => 7+x > 0

=> x > -7

c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )

=> x < 1 ; x > -2

=> -2 < x < 1

Tk mk nha

Bài 3: 

\(\Leftrightarrow3^{2x+6}=3\)

=>2x+6=1

=>2x=-5

hay x=-5/2

a)  (x+3)(x+5)=0

=>x+3=0 hoặc x+5=0

=>x=-3 hoặc -5

b) (x-1).5-1=0

=>5x-5-1=0

=>5x-6=0

=>5x=6

=>x=6/5

c) 

làm câu c,d,b và E đi bạn

OK bạn

b,2x.(x-5)-x.(3+2x)=26

2x² - 10x - 3x - 2x² = 26

-13x = 26

x = -2

c, (x+7)²-x.(x-3)=12

x² +14x +49 - x² + 3x = 12

17x + 49 = 12

17x = - 37

x = \(\dfrac{-37}{17}\)

d, 9( x -2018) - x+ 2018 =0

9( x -2018) - (x -2018) = 0

( 9-1)(x -2018) = 0

8( x -2018) = 0

x -2018 = 0

x = 2018

a: =>2x+10-x^2-5=0

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

=>\(x\in\left\{1+\sqrt{6};1-\sqrt{6}\right\}\)

e: =>4x^2+4x+9x^2-4=15

=>13x^2+4x-19=0

=>\(x\in\left\{\dfrac{-2+\sqrt{251}}{13};\dfrac{-2-\sqrt{251}}{13}\right\}\)

a ) 

\(5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Rightarrow\left(5x+7\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)

Vậy ...

b ) 

\(x^{2017}=x^{2018}\)

\(\Rightarrow x^{2017}-x^{2018}=0\)

\(\Rightarrow x^{2017}\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^{2017}=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy ...

c ) 

\(2x^2=x\)

\(\Rightarrow2x^2:x=1\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy ...

e ) 

\(x^5=x^4\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)( làm tương tự như phần b )  

~ 

d ) 

\(x^3+x^2=0\)

\(\Rightarrow x^2\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

Vậy ...

chẳng có đề bài biết làm ntn

a, Vì -2018 khác 0

=> x-11=0

=> x=11

b, Vì -2018 < 0

=> x+13 > 0

=> x > -13

c, Vì 2018 > 0 => 2x-10 > 0

=> 2x > 10

=> x > 5

d, => x-3=0 hoặc 3x-9=0

=> x=3

e, Vì x-1 < x+5 

=> x-1 < 0 và x+5 > 0

=> x < 1 và x > -5

=> -5 < x < 1

Tk mk nha

a) x( x + 2018 ) - 2x - 4036 = 0 

<=> x( x + 2018 ) - 2( x + 2018 ) = 0

<=> ( x + 2018 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)

b) x + 5 = 2( x + 5 )²

<=> x + 5 = 2( x² + 10x + 25 )

<=> x + 5 = 2x² + 20x + 50

<=> 2x² + 20x + 50 - x - 5 = 0

<=> 2x² + 19x + 45 = 0

<=> 2x² + 10x + 9x + 45 = 0

<=> 2x( x + 5 ) + 9( x + 5 ) = 0

<=> ( x + 5 )( 2x + 9 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)

c) ( x² + 1 )( 2x - 1 ) + 2x = 1

<=> 2x³ - x² + 4x - 1 - 1 = 0

<=> 2x³ - x² + 4x - 2 = 0

<=> x²( 2x - 1 ) + 2( 2x - 1 ) = 0

<=> ( 2x - 1 )( x² + 2 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x² + 2 ≥ 2 > 0 ∀ x )

d) \(\frac{x}{3}-\frac{x^2}{4}=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)

\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)

\(\Leftrightarrow4x-3x^2=0\)

\(\Leftrightarrow x\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)