a, \(\left(x+2\right)^2-5=4\Rightarrow\left(x+2\right)^2=5+4\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\Rightarrow x+2=3\Rightarrow x=3-2=1\)
Vậy x = 1
b, \(\left|1-x\right|+2=-1\Rightarrow\left|1-x\right|=-1+\left(-2\right)\)
\(\Rightarrow\left|1-x\right|=-3\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
c, \(x^2=4x\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[\begin{matrix}x=0\\x=0+4=4\end{matrix}\right.\)
Vậy x = 0 hoặc x = 4
a) \(\left(x+2\right)^2-5=4\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow x+2=3\) hoặc x + 2 = -3
+) \(x+2=3\Rightarrow x=1\)
+) \(x+2=-3\Rightarrow x=-5\)
Vậy \(x\in\left\{1;-5\right\}\)
b) \(\left|1-x\right|+2=-1\)
\(\Rightarrow\left|1-x\right|=-3\)
Mà \(\left|1-x\right|\ge0\)
\(\Rightarrow x\) không có giá trị thỏa mãn
Vậy x không có giá trị thỏa mãn
c) \(x^2=4x\)
\(\Rightarrow x^2-4x=4x-4x\)
\(\Rightarrow x^2-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4\right\}\)
a) \(\left(x+2\right)^2-5=4\)
\(\Leftrightarrow\left(x+2\right)^2=5+4=9\)
\(\Leftrightarrow\left[\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy x = 1 hoặc x = -5
b) \(\left|1-x\right|+2=-1\)
Vì \(\left|1-x\right|\ge0\)
\(\Rightarrow\left|1-x\right|+2\ge2\)
\(\Rightarrow x\in\varnothing\)
c) \(x^2=4x\)
\(\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-4=0\Leftrightarrow x=4\end{matrix}\right.\)
Vậy x = 0 hoặc x = 4
a) Ta có \(\left(x+2\right)^2-5=4\)
\(\Leftrightarrow\left(x+2\right)^2=4+5=9\)
\(\Leftrightarrow x+2=\pm3\)
Với \(x+2=3\Rightarrow x=3-2=1\)
Với \(x+2=-3\Rightarrow x=-3-2=-5\)
Vậy x=1 ;hoặc x=-5
b) ta có \(\left|1-x\right|+2=-1\)
\(\Leftrightarrow\left|1-x\right|=-1-2=-3\) (vô lý)
Do đó không có giá trị nào của x để |1-x|+2=-1
c)ta có \(x^2=4x\)
\(\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\Rightarrow x=4\end{matrix}\right.\)
Vậy x=4 hoặc x=0
a) (x+2)2 - 5 = 4
(x+2)2 = 4+5
(x+2)2 = 9
(x+2)2 = 32
\(\Rightarrow x+2=3\)
x = 3 - 2
x = 1
b) \(\left|1-x\right|+2=-1\)
\(\left|1-x\right|=-1-2\)
\(\left|1-x\right|=-3\)
\(\Rightarrow\left[\begin{matrix}1-x=-3\\1-x=3\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1-\left(-3\right)\\x=1-3\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
c)
\(^{^{^{ }}}x^2=4x\)
\(\Rightarrow x^2-4x=4x-4x\)
\(\Rightarrow x^2-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4\right\}\)
mk làm lại phần b
\(\left|1-x\right|+2=-1\)
\(\left|1-x\right|=-1-2\)
\(\left|1-x\right|=-3\)
mà \(\left|1-x\right|\ge0\)
Vậy \(x\in\varnothing\)
a, \(\left(x+2\right)^2-5=4\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=1\)
b, \(\left|1-x\right|+2=-1\)
\(\Rightarrow\left|1-x\right|=-3\)
Không có giá trị của x thỏa mãn.
c, \(x^2=4x\)
\(\Rightarrow x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, x = 0 hoặc x = 4.
lm ơn giải chi tiết , cụ thể
