Question

Tìm x:

a) 4x+7 chia hết cho x-1

b) x-5 chia hết cho x-6

Các bn jup mk trình bày cả bài luôn nha. Thanks😀

Answer 1

\(4x+7⋮x-1\)

\(\Rightarrow4x-4+11⋮x-1\)

\(\Rightarrow4\left(x-1\right)+11⋮x-1\)

      \(4\left(x-1\right)⋮x-1\)

\(\Rightarrow11⋮x-1\)

\(\Rightarrow x-1\inƯ\left(11\right)=\left\{-1;1;-11;11\right\}\)

\(\Rightarrow x\in\left\{0;2;10;12\right\}\)

vậy_

\(x-5⋮x-6\)

\(\Rightarrow x-6+1⋮x-6\)

     \(x-6⋮x-6\)

\(\Rightarrow1⋮x-6\)

\(\Rightarrow x-6\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow x\in\left\{5;7\right\}\)

vậy_

Answer 2

4x+7⋮x−1

⇒4x−4+11⋮x−1

⇒4(x−1)+11⋮x−1

      4(x−1)⋮x−1

⇒11⋮x−1

⇒x−1∈Ư(11)={−1;1;−11;11}

⇒x∈{0;2;10;12}

vậy_

x−5⋮x−6

⇒x−6+1⋮x−6

     x−6⋮x−6

⇒1⋮x−6

⇒x−6∈Ư(1)={−1;1}

⇒x∈{5;7}

vậy ....................