Bài 1:
a, \(3x-5⋮2x+1\)
\(\Rightarrow2\left(3x-5\right)-3\left(2x+1\right)⋮2x+1\)
\(\Rightarrow6x-10-6x-3⋮2x+1\)
\(\Rightarrow7⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow2x\in\left\{0;-2;6;-8\right\}\)
\(\Rightarrow x\in\left\{0;-1;3;-4\right\}\)
b, \(2x-3⋮x+1\)
\(\Rightarrow2x-3-2\left(x+1\right)⋮x+1\)
\(\Rightarrow2x-3-2x-2⋮x+1\)
\(\Rightarrow1⋮x+1\)
\(\Rightarrow x+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow x\in\left\{0;-2\right\}\)
c, \(3x+2⋮2x-1\)
\(\Rightarrow2\left(3x+2\right)-3\left(2x-1\right)⋮2x-1\)
\(\Rightarrow6x+4-6x+3⋮2x-1\)
\(\Rightarrow7⋮2x-1\)
\(\Rightarrow2x-1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow2x\in\left\{2;0;8;-6\right\}\)
\(\Rightarrow x\in\left\{1;0;4;-3\right\}\)
d, \(2x-1⋮x+3\)
\(\Rightarrow2x-1-2\left(x+3\right)⋮x+3\)
\(\Rightarrow2x-1-2x-6⋮x+3\)
\(\Rightarrow-5⋮x+3\)
\(\Rightarrow x+3\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow x\in\left\{-2;-4;2;-8\right\}\)
Bài 2:
\(\left|x-1\right|\le2\)
\(\Rightarrow-2\le x-1\le2\)
\(\Rightarrow-2+1\le x-1+1\le2+1\)
\(\Rightarrow-1\le x\le3\)
=> x = {-1;0;1;2;3}
* Trả lời:
Bài 2:
\(\left|x-1\right|\le2\)
\(\Rightarrow x-1\le2\) hoặc \(x-1\le-2\)
\(\Rightarrow x\le3\) | \(x\le-1\)
\(\Rightarrow x\inƯ\left\{3\right\}\) | \(x\inƯ\left\{-1\right\}\)
\(\Rightarrow x\in\left\{3;-3;1;-1\right\}\) | \(x\in\left\{-1;1\right\}\)
Vậy \(x\in\left\{3;-3;1;-1\right\}\)
Bài 1:a, 3x−5⋮2x+13x−5⋮2x+1
⇒2(3x−5)−3(2x+1)⋮2x+1⇒2(3x−5)−3(2x+1)⋮2x+1
⇒6x−10−6x−3⋮2x+1⇒6x−10−6x−3⋮2x+1
⇒7⋮2x+1⇒7⋮2x+1
⇒2x+1∈Ư(7)={±1;±7}⇒2x+1∈Ư(7)={±1;±7}
⇒2x∈{0;−2;6;−8}⇒2x∈{0;−2;6;−8}
⇒x∈{0;−1;3;−4}⇒x∈{0;−1;3;−4}
b, 2x−3⋮x+12x−3⋮x+1
⇒2x−3−2(x+1)⋮x+1⇒2x−3−2(x+1)⋮x+1
⇒2x−3−2x−2⋮x+1⇒2x−3−2x−2⋮x+1
⇒1⋮x+1⇒1⋮x+1
⇒x+1∈Ư(1)={±1}⇒x+1∈Ư(1)={±1}
⇒x∈{0;−2}⇒x∈{0;−2}
c, 3x+2⋮2x−13x+2⋮2x−1
⇒2(3x+2)−3(2x−1)⋮2x−1⇒2(3x+2)−3(2x−1)⋮2x−1
⇒6x+4−6x+3⋮2x−1⇒6x+4−6x+3⋮2x−1
⇒7⋮2x−1⇒7⋮2x−1
⇒2x−1∈Ư(7)={±1;±7}⇒2x−1∈Ư(7)={±1;±7}
⇒2x∈{2;0;8;−6}⇒2x∈{2;0;8;−6}
⇒x∈{1;0;4;−3}⇒x∈{1;0;4;−3}
d, 2x−1⋮x+32x−1⋮x+3
⇒2x−1−2(x+3)⋮x+3⇒2x−1−2(x+3)⋮x+3
⇒2x−1−2x−6⋮x+3⇒2x−1−2x−6⋮x+3
⇒−5⋮x+3⇒−5⋮x+3
⇒x+3∈Ư(−5)={±1;±5}⇒x+3∈Ư(−5)={±1;±5}
⇒x∈{−2;−4;2;−8}⇒x∈{−2;−4;2;−8}
bài 2 |x−1|≤2|x−1|≤2
⇒x−1≤2⇒x−1≤2 hoặc x−1≤−2x−1≤−2
⇒x≤3⇒x≤3 | x≤−1x≤−1
⇒x∈Ư{3}⇒x∈Ư{3} | x∈Ư{−1}x∈Ư{−1}
⇒x∈{3;−3;1;−1}⇒x∈{3;−3;1;−1} | x∈{−1;1}x∈{−1;1}
Vậy x∈{3;−3;1;−1}
