b: \(x\in\left\{4;-4\right\}\)

\(a,x^2=16\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ b,\left(2x-1\right)^3=-27\\ \Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\\ \Rightarrow2x-1=-3\\ \Rightarrow2x=-2\\ \Rightarrow x=-1\)

-Chào :)

(2.x+1)³ = 27

(2.x+1)³ = 3³

2.x     = 3 -1

2x      = 2

x        = 2:2

x        = 1

- Học tốt ='>

 (2.+x+1)mu3=27

(2.x+1)mu3=3³

2.x+1=3

2.x=3-1

2.x=2

x=1

( 2x + 1 )\(^3\)= 27

( 2x + 1 )\(^3\)= 3\(^3\)

  2x + 1   =  3

    2x     =  2

      x = 1

 Tk cho mình nhé.

(x-1)³=27

(x-1)³=3³

^x-1=3

^=>x=3+1=4

A=(x²-x+1)2
Có \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>=\frac{3}{4}\)
=>\(A>=\left(\frac{3}{4}\right)^2=\frac{9}{16}\)
MinA=9/16 <=> x=1/2

2x = 7+1

2x = 8

x = 8:2

x =4 

k cho mk nha mn

2x = 8 nha

2x - 1 = 7

2x      = 7 + 1

2x      = 8

  x      = 8 : 2

  x      = 4

Vậy x = 4

1.a)x+7=-5-14

x+7=-19

x=-19-7

x=-26

Vậy x=-26

b)3x-4=(-2)³-11

3x-8=-11

3x=-11+8

3x=-3

x=-3:3

X=-1

Vậy x=-1

c)11+(4x-11)=-9-(-15)

11+(4x-11)=-9+15

11+(4x-11)=6

4x-11=6-11

4x=6

x=\(\dfrac{6}{4}\)

Vậy x=\(\dfrac{6}{4}\)

d)\(\left|2x-1\right|=5\)

⇒2x-1=\(\pm5\)

+Nếu 2x-1=5

2x=6

x=6:2

x=3

Vậy x=3

+Nếu 2x-1=-5

2x=-5+1

2x=-4

x=-4:2

x=-2

Vậy xϵ{3;-2}

e)\(\left|x-7\right|+3=25\)

\(\left|x-7\right|=22\)

⇒x-7=\(\pm22\)

+Nếu x-7=22

x=22+7=29

+Nếu x-7=-22

x=-22+7

x=-15

Vậy xϵ{29;-15}

=2x³-3x-5x³-x²+x²

=-3x³-3x

=-(3x³+3x)

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

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