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: \(x^2-5x=0\)

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

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

|x+1|+|3x-1|+|x-1|=3

=} vs cả trong dấu giá trị tuyệt đối >0 thì=}

x+1+3x-1+x-1=3{=}5x=4{=}x=4/5

=}vs cả trong giá trị tuyệt đối <0 thì=}

x+1+3x-1+x-1=-3{=}5x=-4{=}x=-4/5

a: \(\Leftrightarrow x^2-x+4x-4+5⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{2;0;6;-4\right\}\)

c: \(\Leftrightarrow4x-5=13k\left(k\in Z\right)\)

=>4k=13k+5

hay \(x=\dfrac{13k+5}{4}\)

`123-x=38`

`=> x= 123-38`

`=>x=85`

Vậy `x=85`

__

`(x+15) -7=33`

`=>x+15=33+7`

`=>x+16= 40`

`=>x=40-16`

`=>x=24`

Vậy `x=24`

__

`3^(x+3) -13=230`

`=> 3^(x+3) = 230+13`

`=>3^(x+3)=243`

`=> 3^(x+3)=3^5`

`=> x+3=5`

`=>x=5-3`

`=>x=2`

Vậy `x=2`

a: \(\Leftrightarrow\left(-x+3\right)\left(x+6\right)=18\)

\(\Leftrightarrow-x^2-6x+3x+18-18=0\)

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

=>x=0 hoặc x=-3

b: \(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2x-\dfrac{4}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{\sqrt{21}}{3}-1;\dfrac{-\sqrt{21}}{3}-1\right\}\)

c: =>x(3x-5)=0

=>x=0 hoặc x=5/3

d: =>(x-2)(x+2)=0

=>x=2 hoặc x=-2

\(\left\{{}\begin{matrix}x+y=2\left(m-1\right)\left(1\right)\\2x-y=m+8\left(2\right)\end{matrix}\right.\) 

Cộng từng vế của (1) và (2) ta được: 

\(3x=3m+6=3\left(m+2\right)\) \(\Leftrightarrow x=m+2\) Thay vào (2) ta được:

\(\Rightarrow2\left(m+2\right)-y=m+8\) \(\Leftrightarrow y=2m+4-m-8=m-4\)

\(\Rightarrow x^2+y^2=\left(m+2\right)^2+\left(m-4\right)^2=m^2+4m+4+m^2-8m+16=2m^2-4m+20=2m^2-4m+2+18=2\left(m^2-2m+1\right)+18=2\left(m-1\right)^2+18\ge18\)

GTNN của \(x^2+y^2=18\Leftrightarrow m=1\)