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a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
\(a)\) \(\left|2x-1\right|+5=12-x\)
\(\Leftrightarrow\)\(\left|2x-1\right|=12-x-5\)
\(\Leftrightarrow\)\(\left|2x-1\right|=7-x\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=7-x\\2x-1=x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}2x+x=7+1\\2x-x=-7+1\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=8\\x=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-6\end{cases}}}\)
Vậy \(x=\frac{8}{3}\) và \(x=-6\)
\(b)\) \(\left|2x+1\right|=x-1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=x-1\\2x+1=1-x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-x=-1-1\\2x+x=1-1\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-2\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)
Vậy \(x=-2\) hoặc \(x=0\)
Chúc bạn học tốt ~
a: =>x-562=20
=>x=582
b: =>562-x+568=0
=>1130-x=0
=>x=1130
c: =>2x-123=283-62=221
=>2x=344
=>x=172
d: =>2x-1=2
=>2x=3
=>x=3/2
Bài giải
a) Ta có 2x + 5 \(⋮\)x + 1 (x \(\inℕ\))
Mà 2x + 5 = (x + 1) + (x + 4)
Suy ra (x + 1) + (x + 4) \(⋮\)x + 1
Vì x + 1 \(⋮\)x + 1
Nên x + 4 \(⋮\)x + 1
Suy ra (x + 1) + 3 \(⋮\)x + 1
Mà x + 1 \(⋮\)x + 1
Nên 3 \(⋮\)x + 1
Suy ra x + 1 \(\in\)Ư (3)
Ư (3) = {1; 3}
Nếu x + 1 = 1 thì ta có
x = 1 - 1
x = 0
Nếu x + 1 = 3 thì ta có
x = 3 - 1
x = 2
Vậy x = 0 hoặc x = 2 thì 2x + 5 \(⋮\)x + 1
\(\Rightarrow2^{x-1}=32=2^5\\ \Rightarrow x-1=5\\ \Rightarrow x=6\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
Hình như bạn thiếu đề , bạn bổ sung đề vào nhé !
#Thiếu kết quả
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