\(x^3-3x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(x^3-3x^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy pt có tập nghiệm S = { 0;3 }

khỉ nghĩ như này..

x³-3x²=0

(=)x² (x-3)=0

(=)x²=0,hoac x-3=0

(=)x=3

a) 1^3 + 2^3 = 1 + 8 = 9 (phải, vì 9 = 3^2)

b) 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36 (phải, vì 36 = 6^2)

c) 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100 (phải, vì 100 = 10^2)

\(\left(2x-1\right)\left(3x+5\right)+\left(-6x^3+5x\right):x\)

\(=6x^2+10x-3x-5-6x^3+5x:x\)

\(=-6x^3+6x^2+12x-5:x\)

\(=-6x^2+6x+12-\dfrac{5}{x}=-6\left(x^2-x-12\right)-\dfrac{5}{x}\)

A = (2\(x\) - 1)(3\(x\) + 5) + (-6\(x\)³ + 5\(x\)): \(x\)

A = 6\(x^2\) + 10\(x\) - 3\(x\) - 5 + \(x\)(- 6\(x^2\) + 5): \(x\)

A = 6\(x^2\) + 7\(x\) - 5 - 6\(x^2\) + 5

A = (6\(x^2\) -  6\(x^2\)) + 7\(x\) - (5 - 5)

A =   7\(x\) 

A = 7x

a) \(\left(x-\frac{1}{2}\right)^3=27\)

=> \(\left(x-\frac{1}{2}\right)^3=3^3\)

=> \(x-\frac{1}{2}=3\)

=> \(x=3+\frac{1}{2}\)

=> \(x=\frac{7}{2}\)

Vậy \(x=\frac{7}{2}.\)

b) \(\left(2x-1\right)^3=-27\)

=> \(\left(2x-1\right)^3=\left(-3\right)^3\)

=> \(2x-1=-3\)

=> \(2x=\left(-3\right)+1\)

=> \(2x=-2\)

=> \(x=\left(-2\right):2\)

=> \(x=-1\)

Vậy \(x=-1.\)

Chúc bạn học tốt!

\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)

\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>x=1

vậy ...

\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)

\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

vậy ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

\(\left[{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=1\\x^2=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

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

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)

\(\left(6x-3^3\right).5^3=3.5^4\)

\(\Rightarrow\left(6x-27\right)=3\left(5^4\div5^3\right)\)

\(\Rightarrow\left(6x-27\right)=3.5\)

\(\Rightarrow6x-27=15\)

\(\Rightarrow6x=42\)

\(\Rightarrow x=7\)

e viết thế này k nhìn đc e