`a)` `2x^2=3x`
`<=>2x^2-3x=0`
`<=>x(2x-3)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\2x=3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `x=0;x=3/2`
`b)` `(x-5)^3=x-5`
`<=>(x-5)^3-x+5=0`
`<=>(x-5)^3-(x-5)=0`
`<=>(x-5).[(x-5)^2-1]=0`
`<=>(x-5).(x^2-10x+25-1)=0`
`<=>(x-5).(x^2-10x+24)=0`
`<=>(x-5).(x^2-6x-4x+24)=0`
`<=>(x-5).[(x^2-6x)-(4x-24)]=0`
`<=>(x-5).[x(x-6)-4(x-6)]=0`
`<=>(x-5)(x-6)(x-4)=0`
`<=>` \(\left[ \begin{array}{l}x-5=0\\x-6=0\\x-4=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=5\\x=6\\x=4\end{array} \right.\)
Vậy `x=5;x=6;x=4`
`c)` `|x-1|+x=1`
`<=>|x-1|=1-x` ĐKXĐ: `x\leq1`
`<=>` \(\left[ \begin{array}{l}x-1=1-x\\x-1=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+1\\x-1=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=2\\x-x=-1+1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=1(\text{thoả mãn})\\0x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\x∈R\end{array} \right.\)
Vậy `x\leq1`
`d)` `|x-5|+x=1`
`<=>|x-5|=1-x` Điều kiện: `x\leq1`
`<=>` \(\left[ \begin{array}{l}x-5=1-x\\x-5=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+5\\x-5=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=6\\x-x=-1+5\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3(\text{loại})\\0x=4(\text{vô lý})\end{array} \right.\)
Vậy ta không tìm được giá trị nào của `x`
