Question

[Ngân Khánh]

Tìm x

a.x²-36=0

b.x³-0.25x=0

c.x²-10x=-25

d.x²2x=-1

e.x³+3x²=-3x-1

Answer 1

[Toru]

a) \(x^2-36=0\)

\(\Rightarrow x^2-6^2=0\)

\(\Rightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

b) \(x^3-0,25x=0\)

\(\Rightarrow x\left(x^2-0,25\right)=0\)

\(\Rightarrow x\left[x^2-\left(0,5\right)^2\right]=0\)

\(\Rightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)

c) \(x^2-10x=-25\)

\(\Rightarrow x^2-10x+25=0\)

\(\Rightarrow x^2-2\cdot x\cdot5+5^2=0\)

\(\Rightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

d) \(x^2-2x=-1\) (sửa đề)

\(\Rightarrow x^2-2x+1=0\)

\(\Rightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

e) \(x^3+3x^2=-3x-1\)

\(\Rightarrow x^3+3x^2+3x+1=0\)

\(\Rightarrow\left(x+1\right)^3=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

\(\text{#}Toru\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: \(x^2-36=0\)

=>\(x^2=36=6^2\)

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

b: \(x^3-0,25x=0\)

=>\(x\left(x^2-0,25\right)=0\)

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

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

c: \(x^2-10x=-25\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5

e: \(x^3+3x^2=-3x-1\)

=>\(x^3+3x^2+3x+1=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1