Đáp án:
`a)x∈{3/2;2}`
`b)x∈{5;-3/2}`
`c)x∈{-4/3;4/3}`
`d)x=`$\sqrt[]{3}$
Giải thích các bước giải:
`a)4(2x-3)=2x(2x-3)`
`⇔4(2x-3)-2x(2x-3)=0`
`⇔(2x-3)(4-2x)=0`
`⇔2(2x-3)(2-x)=0`
`⇔(2x-3)(2-x)=0`
`⇔`$\left[\begin{matrix} 2x-3=0\\ 2-x=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\dfrac{3}{2}\\ x=2\end{matrix}\right.$
Vậy `x∈{3/2;2}`
`b)2x(x-5)=3(5-x)`
`⇔2x(x-5)-3(5-x)=0`
`⇔2x(x-5)+3(x-5)=0`
`⇔(x-5)(2x+3)=0`
`⇔`$\left[\begin{matrix} x-5=0\\ 2x+3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=5\\x=-\dfrac{3}{2}\end{matrix}\right.$
Vậy `x∈{5;-3/2}`
`c)9x²-16=0`
`⇔(3x)²-4²=0`
`⇔(3x+4)(3x-4)=0`
`⇔`$\left[\begin{matrix} 3x+4=0\\ 3x-4=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 3x=-4\\ 3x=4\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-\dfrac{4}{3}\\ x=\dfrac{4}{3}\end{matrix}\right.$
Vậy `x∈{-4/3;4/3}`
`d)x^3-3`$\sqrt[]{3}$`x^2+9x-3`$\sqrt[]{3}$`=0`
`⇔x^3-3.x^2.`$\sqrt[]{3}$`+3.x.(`$\sqrt[]{3}$ `)^2-(`$\sqrt[]{3}$ `)^3=0`
`⇔(x-`$\sqrt[]{3}$ `)^3=0`
`⇔x-`$\sqrt[]{3}$`=0`
`⇔x=`$\sqrt[]{3}$
Vậy `x=`$\sqrt[]{3}$
