Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
a. x=0
b.x=1,7
c.x=5,3
G.X=7
h.x=6
Mk làm vậy thôi
hok tốt
Professor minhmama
a;\(x\left(x+0\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+0=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0\end{cases}}}\)
\(b,\left(x-1\right)\left(7-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}}\)
\(c,\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
\(d,\left(x+5\right)+\left(x-9\right)=13\)
\(\Rightarrow x+5+x-9=13\)
\(\Rightarrow2x=17\)
\(\Rightarrow x=\frac{17}{2}\)
\(e;\left(4+x\right)+\left(x-7\right)=x+2\)
\(\Rightarrow4+x+x-7=x+2\)
\(\Rightarrow x=5\)
\(f,\left(3x+5\right)-\left(2x-7\right)=4-x\)
\(\Rightarrow3x+5-2x+7=4-x\)
\(\Rightarrow2x=-8\Rightarrow x=-4\)
\(g,\left(x-1\right)^2=36\)
\(\Rightarrow\left(x-1\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=5\end{cases}}}\)
\(h,\left(3-x\right)^3=-27\)
\(\Rightarrow\left(3-x\right)^3=\left(-3\right)^3\)
\(\Rightarrow3-x=-3\)
\(\Rightarrow x=6\)
a) Ta có: \(x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
Vậy: x∈{0;-7}
b) Ta có: \(\left(x+12\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy: x∈{-12;3}
c) Ta có: \(\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy: x∈{3;5}
d) Ta có: \(x\left(2+x\right)\left(7-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy: x∈{-2;0;7}
e) Ta có: \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy: x∈{-2;1;3}
g) Ta có: \(\left(x-5\right)\left(x^2-81\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)
Vậy: x∈{-9;5;9}
h) Ta có: \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
hay x=-3
Vậy: x=-3