Tìm x
( x + 2 )2 - (x-2).(x+2)=0
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\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
x2.(x2 + 4) - x2 - 4=0
⇒ x2.(x2 + 4) - (x2 + 4) =0
⇒ (x2 + 4) .(x2 - 1) = 0
\(\Rightarrow\left[{}\begin{matrix}x^2+4=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=-4\\x^2=1\end{matrix}\right.\)(loại do x2 ≥ 0) \(\Rightarrow x=\pm1\)
\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)
\(\left(x+2\right)4=0\)
\(x+2=0\)
\(x=-2\)
a) \(x^3=x^5\)
=> \(x^3-x^5=0\)
=> \(x^3\left(1-x^2\right)=0\)
=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(4x\left(x+1\right)=x+1\)
=> \(4x^2+4x-x-1=0\)
=> \(4x\left(x+1\right)-1\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)
c) \(x\left(x-1\right)-2\left(1-x\right)=0\)
=> \(x\left(x-1\right)-\left[-2\left(x+1\right)\right]=0\)
=> \(x\left(x-1\right)+2\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
d) Kết quả ?
e) \(\left(x-3\right)^2+3-x=0\)
=> \(x^2-6x+9+3-x=0\)
=> \(x^2-7x+12=0\)
=> \(x^2-3x-4x+12=0\)
=> \(x\left(x-3\right)-4\left(x-3\right)=0\)
=> (x - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
f) Tương tự
\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)
b) \(x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=\left(0\right)\Leftrightarrow x=1,x=2\)
\(\left(\left|x+2\right|^2\right)\left(\left|2-x\right|^{1871}\right)=0\)
\(\left(x+2\right)^2\left(2-x\right)^{1871}=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+2\right)^2=0\\\left(2-x\right)^{1871}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x+2=0\\2-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
Ta có: \(\left|x+2\right|^2.\left|2-x\right|^{1871}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left|x+2\right|^2=0\\\left|2-x\right|^{1871}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+2=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)
Vậy x = -2 hoặc x = 2
X2(x+2)+4(x+2)=0
=>(x2+4)(x+2)=0
=>x2+4=0 hoặc x+2=0
=>x2=-4 hoặc x=-2
Mà x2 phải ra kết quả là số dương
suy ra x=-2
\(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+4\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-4\\x=-2\end{cases}}}\)
mà \(x^2\ge0\Rightarrow x=-2\)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
\(\Leftrightarrow\)\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\)\(\left(x+2\right)\left(\left(x+2\right)-\left(x-2\right)\right)=0\)
\(\Leftrightarrow\) \(\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow\) \(\left(x+2\right).4=0\)
\(\Leftrightarrow\) \(x+2=0\Leftrightarrow x=-2\)