Bằng cách nhân chéo ta có đẳng thức :
( x + 2 ) ( x - 2 ) = 5 = 5 . 1 = 1 . 5 = ( -1 ) . ( -5 ) = ( -5 ) . ( -1 )
Ta có bảng :
| x + 2 | 1 | 5 | -1 | -5 |
| x - 2 | 5 | 1 | -5 | -1 |
| x | -1 | 3 | -3 | -7 |
| x | 7 | 3 | -3 | 1 |
Vậy x = 3 hoặc -3
\(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\Rightarrow\left(x+2\right).\left(x-2\right)=5\)
\(x-4=5\) ( do (x+2).(x-2) = x2 - 2x + 2x - 4, bn tự phân tích ra nha)
x = 9
\(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\Leftrightarrow\left(x+2\right)\times\left(x-2\right)=1\times5\)
\(\Leftrightarrow\left(x+2\right)\times\left(x-2\right)=5\)
bạn giải tiếp nha!
k tui nha!
\(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\Rightarrow\left(x+2\right).\left(x-2\right)=5\)
Th1 :
\(\left(x+2\right).\left(x-2\right)=-1.\left(-5\right)\Leftrightarrow\hept{\begin{cases}x+2=-1\Leftrightarrow x=-3\\x-2=-5\Leftrightarrow x=-3\end{cases}\left(tm\right)}\)
Th2 :
\(\left(x+2\right).\left(x-2\right)=-5.\left(-1\right)\Leftrightarrow\hept{\begin{cases}x+2=-5\Leftrightarrow x=-7\\x-2=-1\Leftrightarrow x=1\end{cases}\left(loại\right)}\)
Th3 :
\(\left(x+2\right).\left(x-2\right)=1.5\Leftrightarrow\hept{\begin{cases}x+1=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=7\end{cases}\left(loại\right)}\)
Th4 :
\(\left(x+2\right).\left(x-2\right)=5.1\Leftrightarrow\hept{\begin{cases}x+2=5\\x-2=1\end{cases}\Leftrightarrow x=3}\left(tm\right)\)
\(\text{Vậy x = -3 hoặc x = 3}\)
xl bn nha! mk sai
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=> x2 - 4 = 5
x2 = 9 = 32 = (-3)2
=> x = 3 hoặc x = - 3
x^2-4=5 => x= cộng trừ 3
\(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)=1\cdot5=5\)
Ta có : 5 = 1 . 5
= ( -1 ) . ( -5 )
Ta có bảng sau
| x+2 | 1 | -1 | 5 | -5 |
| x | -1 | -3 | 3 | -7 |
| x-2 | 5 | -5 | 1 | -1 |
| x | 7 | -3 | 3 | 1 |
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
\(\frac{x+2}{5}=\frac{1}{x-2}\)\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(2x-4\right)=5\)\(\Leftrightarrow x^2-2x+2x-4=5\)
\(\Leftrightarrow x^2-4=5\)\(\Leftrightarrow x^2=9\)\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
Vậy \(x=-3\)hoặc \(x=3\)
\(\frac{x+2}{5}=\frac{1}{x-2}\)
\(\left(x+2\right)\left(x-2\right)=5\)
\(x^2-4=5\)
\(x^2=9\)
\(x=\pm3\)
Tìm x:
\(\frac{x+2}{5}=\frac{1}{x-2}\)
ĐK :\(x\ne2\)
\(QĐKM:\Rightarrow\left(x+2\right).\left(x-2\right)=5\)
\(\Leftrightarrow x^2-4=5\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\left(tmđk\right)\)
\(S=\left\{3;-3\right\}\)
