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5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
\(\left|2\right|+\left|x+1\right|=3x-2006\) \(\Leftrightarrow2+\left|x+1\right|=3x-2006\)
\(\Leftrightarrow2008+\left|x+1\right|=3x\left(1\right).\)
Vì \(\left|x+1\right|\ge0\) với mọi x nên \(2008+\left|x+1\right|>0.\)
\(\Rightarrow3x>0\Rightarrow x>0\Rightarrow x+1>0\Rightarrow\left|x+1\right|=x+1\)(theo tính chất của giá trị tuyệt đối).
Do đó \(\left(1\right)\Leftrightarrow2008+x+1=3x\)
\(\Leftrightarrow x+2009=3x\)
\(\Leftrightarrow2x=2009\)
\(\Leftrightarrow x=\frac{2009}{2}\)(thoả mãn x > 0).
Vậy \(x=\frac{2009}{2}.\)
Chúc bạn học tốt!
\(\left|2\right|+\left|x+1\right|=3x-2006\)
\(\Leftrightarrow\left|x+1\right|=3x-2008\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=3x-2008\\x+1=2008-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2019}{2}\\x=\frac{2017}{4}\end{cases}}\)
[(3x + 8);2]-6=x
<=>(3x+8):2=x +6
<=>3x + 8=2x + 12
<=>3x - 2x=12-8
<=>x=4
Vậy x=4
~~~~Hok tốt~~~~
[(3x+8):2]-6 = x
<=>(3x+8):2 = x+6
<=> 3x + 8 = 2. (x+6)
<=> 3x + 8 = 2x + 12
<=> 3x - 2x = 12 - 8
<=> x = 4
Vậy x = 4
a) x - 14 = 3x + 18
x - 3x = 18 + 14
-2x = 32
x = 32/2
x = 16
b) 2 ( x - 5) - 3 ( x - 4 ) = -6 + 15 (-3)
2x - 10 -3x + 12 = -6 - 45
2x - 3x = -6 - 45 + 10 - 12
-x = -53
x = 53
hok tốt!!
\(a,\) \(x-14=3x+18\)
\(\Leftrightarrow\) \(-14-18=3x-x\)
\(\Leftrightarrow\) \(-32=2x\)
\(\Leftrightarrow\) \(x=\left(-32\right)\div2\)
\(\Leftrightarrow\) \(x=-16\)
Vậy \(x=-16\)
\(b,\) \(2\left(x-5\right)-3\left(x-4\right)=-6+15\left(-3\right)\)
\(\Leftrightarrow\) \(2x-10-3x+12=-6-45\)
\(\Leftrightarrow\) \(2x-3x=-6-45-12+10\)
\(\Leftrightarrow\) \(-x=-53\)
\(\Leftrightarrow\) \(x=53\)
Vậy \(x=53\)
Hok tốt !
a, 3(x+3)-2(x-5)=11
=> 3x+9-2x+10=11
=> 3x-2x=11-10-9
=> x=-8
Vậy.........
b, 14-4|x|=-6
=> -4|x|=8
=> |x|=-2(VL vì trị tuyệt đối luôn lớn hơn hoặc = 0)
Vậy......
b) 42x=39.42-37.42
42x=42(39-37)
42x=42.2
Vì 42x=42.2
nên x=2
Vậy x=2
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62
Giải:
a) \(4\left(x+1\right)-\left(3x+1\right)=14\)
\(\Leftrightarrow4x+4-3x-1=14\)
\(\Leftrightarrow x+3=14\)
\(\Leftrightarrow x=14-3=11\)
Vậy ...
b) \(-2\left(x-1\right)-\left(2-3x\right)=2\left(x+1\right)-4\)
\(\Leftrightarrow-2x+2-2+3x=2x+2-4\)
\(\Leftrightarrow x=2x-2\)
\(\Leftrightarrow x-2x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)
Vậy ...
c) \(-\left(x-3\right)+\left(4x-1\right)-14=2\left(x-2\right)-5\)
\(\Leftrightarrow-x+3+4x-1-14=2x-4-5\)
\(\Leftrightarrow3x-15=2x-9\)
\(\Leftrightarrow3x-2x=-9+15\)
\(\Leftrightarrow x=6\)
Vậy ...
Ta bảng xét dấu:
*) Nếu \(x\le-2\)
\(\Rightarrow-x-2-3x=4\)
\(\Rightarrow-4x-2=4\)
\(\Rightarrow-4x=6\)
\(\Rightarrow x=\frac{-6}{4}\)(ko thỏa mãn \(x\le-2\))
*) Nếu \(x>-2\)
\(\Rightarrow x+2-3x=4\)
\(\Rightarrow-2x+2=4\)
\(\Rightarrow-2x=2\)
\(\Rightarrow x=-1\)(thỏa mãn x>-2)
Vậy x=-1