b: =>12-x=-7

hay x=19

Bài 4:

a: |x-2|=0

=>x-2=0

=>x=2

b: |x+3|=1

=>\(\left[\begin{array}{l}x+3=1\\ x+3=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1-3=-2\\ x=-1-3=-4\end{array}\right.\)

c: \(-3\left|4-x\right|=-9\)

=>|4-x|=3

=>|x-4|=3

=>\(\left[\begin{array}{l}x-4=3\\ x-4=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=1\end{array}\right.\)

d: |2x+1|=-2

mà -2<0<=|2x+1|

nên x∈∅

Bài 5:

a: \(\left(x+3\right)^2=36\)

=>\(\left[\begin{array}{l}x+3=6\\ x+3=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6-3=3\\ x=-6-3=-9\end{array}\right.\)

b: \(\left(x+5\right)^2=100\)

=>\(\left[\begin{array}{l}x+5=10\\ x+5=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10-5=5\\ x=-10-5=-15\end{array}\right.\)

c: \(\left(2x-4\right)^2=0\)

=>2x-4=0

=>2x=4

=>x=2

d: \(\left(x-1\right)^3=27\)

=>x-1=3

=>x=4

\(a,\Rightarrow x=30-18=12\\ b,\Rightarrow x+6=45:5=9\\ \Rightarrow x=9-6=3\\ c,\Rightarrow38-3x=4^2=16\\ \Rightarrow3x=38-16=22\\ \Rightarrow x=\dfrac{22}{3}\)

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4