Timf x \(\varepsilon\)Z
x.(x+3)=0
(x-2).(5-x)=0
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\(\text{a) 3.(x-2)+x.(x-2)=0}\)
\(\Leftrightarrow\)\(\text{(x-2)(3+x)=0}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\3+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-3}\)
\(b,4x.\left(x-2\right)-x+2\)=0
\(\Leftrightarrow4x.\left(x-2\right)-\left(x-2\right)\)=0
\(\Leftrightarrow\left(x-2\right)\left(4x-1\right)\)=0
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{4}\end{array}\right.\)
Vậy x=2 hoặc \(x=\frac{1}{4}\)
\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)
Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
a) pt
<=> (x - 5)(x + 5) - (x - 5) = 0
<=> (x - 5)(x + 4) = 0
<=> x - 5 = 0 hoặc x + 4 = 0
<=> x = 5 hoặc x = -4
b) pt
<=> (2x - 1)(2x - 1 - 2x - 1) = 0
<=> (2x - 1).(-2)=0
<=> 2x - 1 = 0
<=> x = 1/2
c) pt
<=> (x - 1)(x + 1)(x^2 + 4) = 0
<=> x - 1 = 0 hoặc x + 1 = 0 hoặc x^2 + 4 = 0
<=> x = 1 hoặc x = -1
\(\Leftrightarrow x^2-2x+1-9x^2+36x-36=0\\ \Leftrightarrow-8x^2+34x-35=0\\ \Leftrightarrow8x^2-34x+35=0\\ \Leftrightarrow8x^2-20x-14x+35=0\\ \Leftrightarrow\left(2x-5\right)\left(4x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{7}{4}\end{matrix}\right.\)
a) \(x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy x=0 hoặc x=-2
b) \(x^2\left(x-5\right)+2\left(x-5\right)=0\)
\(\Rightarrow x-5=0\Rightarrow x=5\)
Vậy x=5
a) x(x-2)=0
=> x=0 hoặc x-2=0
=> x=0 hoặc x=0+2
=> x=0 hoặc x = 2
\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
x.(x+3) = 0
<=> x=0 hoặc x+3=0
<=> x=0 hoặc x=-3
(x-2).(5-x)=0
<=> x-2=0 hoặc 5-x=0
<=> x=2 hoặc x=5
a, x=0
hoặc x+3=0=>x=-3
b,x-2=0=>x=2
hoặc 5-x=0=>x=5