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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 \(x\in\left\{0;-2;1;-3\right\}\)
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......
\(a,2x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\forall Z\\x=1\end{cases}}}\)
\(b,x\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
\(c;\left(x+1\right)+\left(x+3\right)+...............+\left(x+99\right)=0\)
\(\Rightarrow\left(x+x+...........+x\right)+\left(1+3+............+99\right)=0\)
\(\Rightarrow50x+2500=0\)
\(\Rightarrow50x=-2500\)
\(\Rightarrow x=-50\)
2/
\(a;\left(x-3\right)\left(2y+1\right)=7\)
\(\Rightarrow\left(x-3\right);\left(2y+1\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Xét bảng
| x-3 | 1 | -1 | 7 | -7 |
| 2y+1 | 7 | -7 | 1 | -1 |
| x | 4 | 2 | 10 | -4 |
| y | 3 | -4 | 0 | -1 |
Vậy...............................
\(b;xy+3x-2y=11\)
\(\Rightarrow x\left(y+3\right)-2y-6=11-6\)
\(\Rightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Rightarrow\left(x-2\right)\left(y+3\right)=5\)
\(\Rightarrow\left(x-2\right);\left(y+3\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét bảng'
| x-2 | 1 | -1 | 5 | -5 |
| y+3 | 5 | -5 | 1 | -1 |
| x | 3 | 1 | 7 | -3 |
| y | 2 | -8 | -2 | -4 |
Vậy................................
\(a,x\left(4-y\right)=3\)
\(\Rightarrow x;4-y\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Tự lập bảng ...
\(b,\left(x-1\right)\left(5-y\right)=7\)
\(Th1:x-1=7\Leftrightarrow x=8\)
\(5-y=1\Leftrightarrow y=4\)
\(Th2:x-1=1\Leftrightarrow x=2\)
\(5-y=7\Leftrightarrow x=-2\)
\(Th3:x-1=-7\Leftrightarrow x=-6\)
\(5-y=-1\Leftrightarrow y=6\)
\(Th4:x-1=-1\Leftrightarrow x=0\)
\(5-y=-7\Leftrightarrow x=12\)
\(c,\left(xy-3\right)\left(x+2\right)=-5\)
\(\Rightarrow xy-3;x+2\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
Tự lập bảng ...
a) Xét \(Ư\left(3\right)=1;3;-1;-3\Leftrightarrow x\left(4-y\right)=3\) có bốn trường hợp
\(TH1:x=1\Leftrightarrow\left(4-y\right)=3\Rightarrow y=4-3=1\)
\(TH2:x=3\Rightarrow\left(4-y\right)=1\Leftrightarrow y=4-1=3\)
\(TH3:x=-1\Rightarrow\left(4-y\right)=-3\Leftrightarrow y=4-\left(-3\right)=7\)
\(TH4:x=-3\Rightarrow\left(4-y\right)=-1\Leftrightarrow y=4-\left(-1\right)=5\)
b) Xét \(Ư\left(7\right)=1;7;-1;-7\Rightarrow\left(x-1\right)\left(5-7\right)\) có bốn trường hợp
\(TH1:x-1=1\Leftrightarrow x=1+1=2\Rightarrow\left(5-y\right)=7\Leftrightarrow v=5-7=-2\)
\(TH2:x-1=7\Leftrightarrow x=7+1=8\Rightarrow\left(5-y\right)=1\Leftrightarrow y=5-1=4\)
\(TH3:x-1=-1\Leftrightarrow x=0\Rightarrow\left(5-y\right)=-7\Leftrightarrow v=12\)
\(TH4:x-1=-7\Leftrightarrow x=-6\Rightarrow\left(5-y\right)=-1\Leftrightarrow y=6\)
Chứng minh tương tự với trường hợp c
Lm câu 2 trc nhé:
\(x-3+x-3=\left(x-3\right)+\left(x-3\right)=2\left(x-3\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
Chỉ lm tắt thôi ạ, hiểu rồi tự trình bày nha~
\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x}{3}-\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5-3}{15}\)
\(\Leftrightarrow4.15=x5-3y\)
\(\Leftrightarrow60=x5-3y\)
\(\Leftrightarrow x5-3y=60\)
tìm x,y như bt nhé
a;\(\left|x-1\right|+\left|3-2\right|=2\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
b;\(\left|x-2\right|+\left|x-3\right|=1\)
\(\Rightarrow\left|x-2\right|+\left|3-x\right|=1\)
Ta có \(\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)
\(\Leftrightarrow\left(x-2\right)\left(3-x\right)\ge0\Leftrightarrow2\le x\le3\)
2/\(M=1-2+3-4+5-6+...........+19-20\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...........+\left(19-20\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+.............+\left(-1\right)\)
\(=\left(-1\right).10\)
a) \(/x-2/+2-x=0\)
\(x-2+2-x=0\)
\(x-2=0\)
\(x=2\)
b) \(/x-3/-3=(-x)\)
\(x-3-3=\left(-x\right)\)
\(x-3=\left(-3\right)\)
\(x=\left(-3\right)+3\)
\(x=0\)
\(x=0\)
mình quên ! mình ghi thêm x =0