`a)(x-4)^2-36=0`
`⇔(x-4)^2-6^2=0`
`⇔(x-4-6).(x-4+6)=0`
`⇔(x-10).(x+2)=0`
\(⇔\left[ \begin{array}{l}x-10=0\\x+2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=10\\x=-2\end{array} \right.\)
Vậy `x∈ {10;-2}`
`b)x.(x-5)-4x+20=0`
`⇔x.(x-5)-(4x-20)=0`
`⇔x.(x-5)-4.(x-5)=0`
`⇔(x-5).(x-4)=0`
\(⇔\left[ \begin{array}{l}x-5=0\\x-4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=5\\x=4\end{array} \right.\)
Vậy `x∈{5;4}`
`c)x^4-2x^3+10x^2-20x=0`
`⇔(x^4-2x^3)+(10x^2-20x)=0`
`⇔x^3.(x-2)+10x.(x-2)=0`
`⇔(x-2).(x^3+10x)=0`
`⇔x.(x-2).(x^2+10)=0`
\(⇔\left[ \begin{array}{l}x=0\\x=2\\x^2+10=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x-2=0\\x^2=-10(vô lý)\end{array} \right.\)
Vậy`x∈{0 ;2}`
`d)(5-3x).4x-15.(3x-5)=0`
`⇔(5-3x).4x+15.(5-3x)=0`
`⇔(5-3x).(4x+15)=0`
\(⇔\left[ \begin{array}{l}5-3x=0\\4x+15=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}5x=3\\4x=-15\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\frac{3}{5}\\x=\frac{-15}{4}\end{array} \right.\)
Vậy `x∈{3/5 ;-15/4}`
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