a. x^2 - 9 = 0 b. x^2 + 1 + 0
c.x^2=2 d. x^2 - 3 = 0
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TA
1
13 tháng 10 2021
c: Ta có: \(x^3+3x^2+3x-7=0\)
\(\Leftrightarrow x+1=2\)
hay x=1
b: Ta có: \(x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
27 tháng 11 2021
\(a,x\left(x-3\right)=x^2-6\\ \Rightarrow x^2-3x-x^2=-6\\ \Rightarrow-3x=-6\\ \Rightarrow x=2\\ b,x^2-7x+12=0\\ \Rightarrow\left(x^2-3x\right)-\left(4x-12\right)=0\\ \Rightarrow x\left(x-3\right)-4\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ d,x^3-25x=0\\ \Rightarrow x\left(x^2-25\right)=0\\ \Rightarrow x\left(x-5\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x0=\\x=5\\x=-5\end{matrix}\right.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
a, (\(x-2\))2 - (2\(x\) + 3)2 = 0
(\(x\) - 2 - 2\(x\) - 3)(\(x\) - 2 + 2\(x\) + 3) = 0
(-\(x\) - 5)(3\(x\) +1) = 0
\(\left[{}\begin{matrix}-x-5=0\\3x+1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\3x=-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\) { -5;- \(\dfrac{1}{3}\)}
b, 9.(2\(x\) + 1)2 - 4.(\(x\) + 1)2 = 0
{3.(2\(x\) + 1) - 2.(\(x\) +1)}{ 3.(2\(x\) +1) + 2.(\(x\) +1)} = 0
(6\(x\) + 3 - 2\(x\) - 2)(6\(x\) + 3 + 2\(x\) + 2) = 0
(4\(x\) + 1)(8\(x\) + 5) =0
\(\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{8}\end{matrix}\right.\)
S = { - \(\dfrac{5}{8}\); \(\dfrac{-1}{4}\)}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
d, \(x^2\)(\(x\) + 1) - \(x\) (\(x+1\)) + \(x\)(\(x\) -1) = 0
\(x\left(x+1\right)\).(\(x\) - 1) + \(x\)(\(x\) -1) = 0
\(x\)(\(x\) -1)(\(x\) + 1 + 1) = 0
\(x\left(x-1\right)\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
S = { -2; 0; 1}
ND
Nguyễn Đức Trí
VIP
18 tháng 7 2023
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
SP
3
17 tháng 9 2018
\(a)\)\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\)\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt ~
\(a,\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^2=-1\left(vô.lí\right)\Rightarrow x\in\varnothing\\ c,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\ d,\Rightarrow x^2=3\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
a) \(\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
b) \(x^2+1=0\)
\(\Rightarrow x^2=-1\left(VLý.do.x^2\ge0\forall x\right)\)
Vậy \(S=\varnothing\)
c) \(\Rightarrow x=\pm\sqrt{2}\)
d) \(\Rightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)