`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`

Đáp án:

1, `2x^2 = 3x`

   `2x^2 - 3x = 0`

   `x(2x - 3)=0`

⇒\(\left[ \begin{array}{l}x=0\\2x - 3 =0\end{array} \right.\) 

⇒\(\left[ \begin{array}{l}x=0\\x=3/2\end{array} \right.\) 

Vậy `x ∈{ 0 ; 3/2}`

2, `(x-5)^3 = x - 5`

   `(x-5)^3 - (x-5) = 0`

   `(x-5).((x-5)^2 - 1)=0`

⇒\(\left[ \begin{array}{l}x - 5  = 0\\(x-5)² - 1 = 0\end{array} \right.\) 

⇒\(\left[ \begin{array}{l}x=5\\x= 4 hoặc x = 6\end{array} \right.\) 

Vậy` x ∈{ 4 ; 5 ; 6}`

3, `| x - 1| + x = 1`

   `|x - 1|         = 1 - x`

⇒\(\left[ \begin{array}{l}x - 1 = 1 - x\\x - 1 = -1 + x\end{array} \right.\) 

⇒\(\left[ \begin{array}{l}x=1\\0x = -2( Vô lí)\end{array} \right.\) 

Vậy `x = 1`

4, `|x-5|+x=1`

   `| x - 5|  = 1 - x`

⇒\(\left[ \begin{array}{l}x - 5 = 1 - x\\x - 5 =-1 + x\end{array} \right.\) 

⇒\(\left[ \begin{array}{l}x=3(Loại)\\0x = 4(Vô lí)\end{array} \right.\) 

Vậy `x ∈ {∅}`