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a) \(10⋮\left(x-1\right)\) (đkxđ \(x\ne1\))
\(\Rightarrow x-1\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)
\(\Rightarrow x\in\left\{0;2;-1;3;-4;6;-9;11\right\}\)
b) \(\left(x+5\right)⋮\left(x-2\right)\left(đkxđ,x\ne2\right)\)
\(\Rightarrow\left(x+5\right)-\left(x-2\right)⋮\left(x-2\right)\)
\(\Rightarrow x+5-x+2⋮\left(x-2\right)\)
\(\Rightarrow7⋮\left(x-2\right)\)
\(\Rightarrow x-2\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{1;3;-5;9\right\}\)
c) \(\left(3x+8\right)⋮\left(x-1\right)\left(đkxd,x\ne1\right)\)
\(\Rightarrow\left(3x+8\right)-3\left(x-1\right)⋮\left(x-1\right)\)
\(\Rightarrow3x+8-3x+3⋮\left(x-1\right)\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1\in\left\{-1;1;-11;11\right\}\)
\(\Rightarrow x\in\left\{0;2;-10;12\right\}\)
a) x∈{0;2;−1;3;−4;6;−9;11}
b) x∈{1;3;−5;9}
c) x ∈ {0;2;−10;12}
a.
\(10⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(10\right)\)
\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)
b.
\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;1;3;9\right\}\)
c.
\(\left(3x+8\right)⋮\left(x-1\right)\)
\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)
\(\Rightarrow3\left(x-1\right)+11⋮x-1\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x=\left\{-10;0;2;12\right\}\)
a) <=> 10 - 2x + 5 = 1 - 3x
<=> -2x + 3x = 1 - 10 - 5
<=> x = -14
b) \(\Leftrightarrow\frac{x-1}{x+1}-\frac{8}{9}=0\)
\(\Leftrightarrow\frac{9\left(x-1\right)-8\left(x+1\right)}{9\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{9x-9-8x-8}{9x+9}=0\)
\(\Leftrightarrow\frac{x-17}{9x+9}=0\)
ĐK: 9x + 9 \(\ne\)0 => x \(\ne\)-1
\(\Leftrightarrow x-17=0\)
\(\Leftrightarrow x=17\)
Ps: Câu b không chắc lắm.
phan b Vu Nhu Mai sai roi phai the nay moi dung
x-1/x+1=8/9
(x-1).9=(x+1).8
9x-9=8x+8
9x-8x=8+9
1x=17
x=17;1
x=17
a) 2x + 1 = 3
2x = 2
x = 1
b) ( 2x - 5 ) + 17 = 6
( 2x - 5 ) = 6 - 17
( 2x - 5 ) = -11
2x = -11 + 5
2x = -6
x = -3
c) 10 - 2 x ( 4 - 3x ) = -4
2 x ( 4 - 3x ) = 14
4 - 3x = 7
3x = -3
x = -1
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) 3x - 5 = -8 - 18
3x-5=-26
3x=-26+5
3x=-21
x=-21:3
x=-7
vậy x=-7
b) /x/ - 10 = -3
|x|=-3+10
|x|=7
=> x=7 hoặc x=-7
vậy x=7
hoặc x=-7
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
| x+1 | 1 | -1 | 5 | -5 |
| y-2 | 5 | -5 | 1 | -1 |
| x | 0 | -2 | 4 | -6 |
| y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
| x-5 | 1 | -1 | 7 | -7 |
| y+4 | -7 | 7 | -1 | 1 |
| x | 6 | 4 | 12 | -2 |
| y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
a) 10 ⋮ (x - 1)
⇒ x - 1 ∈ Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}
⇒ x ∈ {-9; -4; -1; 0; 2; 3; 6; 11}
b) x + 5 = x - 2 + 7
Để (x - 5) ⋮ (x - 2) thì 7 ⋮ (x - 2)
⇒ x - 2 ∈ Ư(7) = {-7; -1; 1; 7}
⇒ x ∈ {-5; 1; 3; 9}
c) 3x + 8 = 3x - 3 + 11
= 3(x - 1) + 11
Để (3x + 8) ⋮ (x - 3) thì 11 ⋮ (x - 3)
⇒ x - 3 ∈ Ư(11) = {-11; -1; 1; 11}
⇒ x ∈ {-8; 2; 4; 14}