a: =>2^4x<2^28

=>4x<28

=>x<7

b: =>5^3x+3<5

=>3x+3<1

=>3x<-2

=>x<-2/3

a) \(16^x< 128^4\)

= (24)^x < (27)⁴

= 2^4x < 2²⁸

= 4x < 28

= x < 7 

Vậy \(x=\left\{0;1;2;3;4;5;6;\right\}\)

\(#Tuyết\)

CÁI NÀY EM KO BIẾT 

EM ĐANG HỌC LỚP 5

a) x^2+3x=0
<=> x(x+3)=0
<=> x+3=0 

---> X=-3
b)x.(x-7).(x+7)=0 

<=>x.(x^2-7^2)=0
<=> X^2-7^2=0
==>x= 7 và x=-7
c) x^3-9x=0
<=> x(x^2-3^2)=0
<=> x^2-3^2=0
~~> x = 3 và x=-3
d) x^2-5x-6=0
<=> x^2-5x-5-1=0
<=> (x^2-1)-(5x-5) =0
<=> x(x-1) - 5(x-1)=0
<=> (x-1)(x-5)=0
~~> x-1 = 0 ~> x=1
~~> x-5=0 ~~> x=5
Vậy x=1 và x=5
 

5x( x - 7 ) - 40( x - 7 ) = 0

<=> ( x - 7 )( 5x - 40 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\5x-40=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

x( x + 5 ) - 7( x + 5 ) = 0

<=> ( x + 5 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)

Ta có: \(5x.\left(x-7\right)-40.\left(x-7\right)=0\)

=> \(\orbr{\begin{cases}5x-40=0\\x-7=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

Ta có: \(x.\left(x+5\right)-7.\left(x+5\right)=0\)

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

=> \(\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)

Mình làm hơi tắt xíu :v ~~~

\(5x.\left(x-7\right)-40.\left(x-7\right)=0\)

\(\left(5x-40\right).\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-40=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=40\\x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=7\end{cases}}}\)

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

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

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

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3