1: (5-x)(x+7)=0
=>\(\left[{}\begin{matrix}5-x=0\\x+7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
2: (8+x)(6-x)=0
=>\(\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
3: 8x(5-x)=0
=>x(5-x)=0
=>\(\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
4: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5: (5x+1)2=121
=>\(\left[{}\begin{matrix}5x+1=11\\5x+1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=10\\5x=-12\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{12}{5}\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
`#3107.101107`
`1)`
`(5 - x)(x + 7) = 0`
TH1: `5 - x = 0 => x = 5`
TH2: `x + 7 = 0 => x = -7`
Vậy, `x \in {-7; 5}`
`2)`
`(8 + x)(6 - x) = 0`
TH1: `8 + x = 0 => x = -8`
TH2: `6 - x = 0 => x = 6`
Vậy, `x \in {-8; 6}`
`3)`
`8x(5 - x) = 0`
TH1: `8x = 0 => x = 0`
TH2: `5 - x = 0 => x = 5`
Vậy, `x \in {0; 5}`
`4)`
`x^2 - 5x = 0`
`=> x*(x - 5) = 0`
TH1: `x = 0`
TH2: `x - 5 = 0 => x = 5`
Vậy, `x \in {0; 5}`
`5)`
`(5x + 1)^2 = 121`
`=> (5x + 1)^2 = (+-11)^2`
TH1: `5x + 1 = 11`
`=> 5x = 11 - 1`
`=> 5x = 10`
`=> x = 2`
TH2: `5x + 1 = -11`
`=> 5x = -11 - 1`
`=> 5x = -12`
`=> x = -12/5`
Mà x là số nguyên `=>` loại
Vậy, `x = 2.`
1.
$(5-x)(x+7)=0$
$\Rightarrow 5-x=0$ hoặc $x+7=0$
$\Rightarrow x=5$ hoặc $x=-7$
2.
$(8+x)(6-x)=0$
$\Rightarrow 8+x=0$ hoặc $6-x=0$
$\Rightarrow x=-8$ hoặc $x=6$
3.
$8x(5-x)=0$
$\Rightarrow 8x=0$ hoặc $5-x=0$
$\Rightarrow x=0$ hoặc $x=5$
4.
$x^2-5x=0$
$\Rightarrow x(x-5)=0$
$\Rightarrow x=0$ hoặc $x-5=0$
$\Rightarrow x=0$ hoặc $x=5$
5.
$(5x+1)^2=121=11^2=(-11)^2$
$\Rightarrow 5x+1=11$ hoặc $5x+1=-11$
Nếu $5x+1=11\Rightarrow x=2$
Nếu $5x+1=-11\Rightarrow x=\frac{-12}{5}$ không là số nguyên (loại)
Vậy $x=2$
