x^3+ 64= 0

x^3 = -64

x^3 = (-4)^3

x=-4

x³ + 64 = 0

x³= 0+64

x³ = 64 

x³= 4³

=> x=4

\(x^3+64=0\)

\(\Rightarrow x^3=-64\)

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

\(\Rightarrow x=-4\)

Vậy \(x=-4\)

\(\left(5x-2\right)^3-64^3=0\Leftrightarrow\left(5x-2\right)^3=64^3\Leftrightarrow5x-2=64\)

\(\Leftrightarrow5x=66\Leftrightarrow x=\dfrac{66}{5}\)

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => 4x² - 12x + 9 - x² - 10x - 25 = 0 

=> 3x² - 22x - 16 = 0

=> (x - 8)(3x + 2) = 0

=> x - 8 = 0 => x = 8

hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3

Vậy x = 8 ; x = -2/3

c/ => x³ - x² - 4x² + 8x - 4 = 0 

=> x³ - 5x² + 8x - 4 = 0 

=> (x - 2)² (x - 1) = 0

=> (x - 2)² = 0 => x - 2 = 0 => x = 2

hoặc x - 1 = 0 => x = 1 

Vậy x = 2 ; x = 1

a) \(x-452=77+48\)

\(=>x-452=125\)

\(=>x=125+452\)

\(=>x=577\)

____

b) \(x+58=64+58\)

\(=>x+58=122\)

\(=>x=122-58\)

\(=>x=64\)

_____

c) \(x-1-2-3-4=0\)

\(=>x-10=0\)

\(=>x=0+10\)

\(=>x=10\)

\(ĐK:x\ge0\\ PT\Leftrightarrow\left(x-\dfrac{3}{4}\right)\left(x^2+\dfrac{3}{4}x+\dfrac{9}{16}\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(n\right)\\\sqrt{x}=3\left(n\right)\\x^2+2\cdot\dfrac{3}{8}x+\dfrac{9}{64}+\dfrac{27}{64}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\\\left(x+\dfrac{3}{8}\right)^2+\dfrac{27}{64}=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\end{matrix}\right.\)

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos

hahaha

\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a)-28+x=-34+(-11)             b)-12(3+x)=0

<=>-28+x=-45                   <=>-36-12x=0

<=>x=-17                            <=>-12x=36

Vậy x=-17                             <=>x=-3

                                         Vậy x=-3

c)(7-x)(-x+2)=0

<=>7-x=0 hoặc -x+2=0

Th1:7-x=0                    Th2:-x+2=0

 <=>x=7                             <=>x=2

           Vậy xϵ{7;2}

d)16x2-64=0

<=>32x=64

<=>x=2

Vậy x=2