`a)x(x-5)=0`
`<=>[(x=0),(x-5=0<=>x=5):}`
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`b)4x(x-7)+(x-7)=0`
`<=>16x^2-28x+x-7=0`
`<=>16x^2-27x-7=0`
`<=>16x^2-2.4x . 27/8+729/64=1177/64`
`<=>(4x-27/8)^2=1177/64`
`<=>|4x-27/8|=\sqrt{1177}/8`
`<=>x=[27+-\sqrt{1177}]/32`
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`c)x^2-4x+3=0`
`<=>x^2-3x-x+3=0`
`<=>(x-3)(x-1)=0`
`<=>[(x=3),(x=1):}`
\(a,x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{0;5\right\}\)
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\(b,4x\left(4x-7\right)+\left(x-7\right)=0\)
\(16x^2-28x+x-7=0\)
\(16x^2-27x-7=0\) (bài này làm ra được nhưng số ra hơi lẻ, bạn xem lại đề)
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\(c,x^2-4x+3=0\)
\(x^2-3x-x+3=0\)
\(x\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}\)
`a, x(x - 5) = 0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
`----------`
`b,4x(4x - 7) + (x - 7) = 0`
`->` đề có sai ko ạ!?
`----------`
`c) x^2 - 4x + 3 = 0`
`<=>x^2-3x-x+3=0`
`<=> x(x-3)-(x-3)=0`
`<=>(x-3)(x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
\(a\)) \(x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(\Rightarrow x\in\left\{0;5\right\}\)
\(b\)) Sửa đề: \(4x\left(x-7\right)+\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\4x=-1\Rightarrow x=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow x\in\left\{7;-\dfrac{1}{4}\right\}\)
\(c\)) \(x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
\(\Rightarrow x\in\left\{3;1\right\}\)
