Đáp án + Giải thích các bước giải:
Ta có :
`a,|2x-7|+2=13`
`→|2x-7|=11`
`→` \(\left[ \begin{array}{l}2x-7=11\\2x-7=-11\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=18\\2x=-4\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=9\\x=-2\end{array} \right.\)
Vậy `x∈{9;-2}`
`b,(2x-1)^2=9`
`→` \(\left[ \begin{array}{l}(2x-1)^2=3^2\\(2x-1)^2=(-3)^2\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x-1=3\\2x-1=-3\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy `x∈{2;-1}`
`----------------`
Ta có :
`3x+8`
`=(3x-3)+11`
`=3(x-1)+11`
Vì `3(x-1)` $\vdots$ `x-1`
Nên để `3x+8` $\vdots$ `x-1`
Thì `11` $\vdots$ `x-1` `(ĐK:x-1\ne0→x\ne1)`
`→x-1∈Ư(11)`
`→x-1∈{±1;±11}`
`→x∈{0;2;-10;12}` ( Thỏa Mãn )
Vậy để `3x+8` $\vdots$ `x-1` thì `x∈{0;2;-10;12}`
