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\(\left(y-6\right)^3-\left(y-6\right)^2=0\)
\(\left(y-6\right)^2.\left[\left(y-6\right)-1\right]=0\)
\(=>\orbr{\begin{cases}y^2-6=0\\y-7=0\end{cases}}\) \(=>\orbr{\begin{cases}y=6\\y=7\end{cases}}\)
Vậy ...
1)\(\left(x+1\right).\left(y-2\right)=0\) \(\left(x,y\inℤ\right)\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
2)\(\left(x-5\right).\left(y-7\right)=1\)
x-5 | 1 | -1 |
y-7 | 1 | -1 |
x | 6 | 4 |
y | 8 | 6 |
3)\(\left(x+4\right).\left(y-2\right)=2\)
x+4 | 1 | 2 | -1 | -2 |
y-2 | 2 | 1 | -2 | -1 |
x | -3 | -2 | -5 | -6 |
y | 4 | 3 | 0 | 1 |
4)\(\left(x-4\right).\left(y+3\right)=-3\)
x-4 | 1 | -1 | 3 | -3 |
y+3 | -3 | 3 | -1 | 1 |
x | 5 | 3 | 7 | 1 |
y | -6 | 0 | -4 | -2 |
5)\(\left(x+3\right).\left(y-6\right)=-4\)
x+3 | -1 | 1 | -4 | 4 | 2 | -2 |
y-6 | 4 | -4 | 1 | -1 | -2 | 2 |
x | -4 | -2 | -7 | 1 | -1 | -5 |
y | 10 | 2 | 7 | 5 | 4 | 8 |
6)\(\left(x-8\right).\left(y+7\right)=5\)
x-8 | 1 | 5 | -1 | -5 |
y+7 | 5 | 1 | -5 | -1 |
x | 9 | 13 | 7 | 3 |
y | -2 | -6 | -12 | -8 |
7)\(\left(x+7\right).\left(y-3\right)=-6\)
x+7 | -1 | 1 | -6 | 6 | -2 | 2 | -3 | 3 |
y-3 | 6 | -6 | 1 | -1 | 3 | -3 | 2 | -2 |
x | -8 | -6 | -13 | -1 | -9 | -5 | -10 | -4 |
y | 9 | -3 | 4 | 2 | 6 | 0 | 5 | 1 |
8)\(\left(x-6\right).\left(y+2\right)=7\)
x-6 | 1 | 7 | -1 | -7 |
y+2 | 7 | 1 | -7 | -1 |
x | 7 | 13 | 5 | -1 |
y | 5 | -1 | -9 | -3 |
ok :)
1/ (x+1)(y+2) =5
Do x;y thuộc N nên x+1 ; y+2 cũng thuộc N
\(TH1:\Leftrightarrow\hept{\begin{cases}x+1=1\\y+2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-1\\y=5-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\y=3\end{cases}}}\\\)
\(TH2:\Leftrightarrow\hept{\begin{cases}x+1=5\\y+2=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5-1\\y=1-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=-1\end{cases}}}\)
x | 0 | 4 |
y | 3 | -1 |
mà x;y\(\in\)N nên x;y=0;3
Các bài khác bạn làm tương tự nha! (vì mk viết rất chậm )
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
1) \(\frac{3}{x}+\frac{y}{3}=\frac{5}{6}\)
\(\Leftrightarrow\frac{3}{x}=\frac{5}{6}-\frac{y}{3}\)
\(\Leftrightarrow\frac{3}{x}=\frac{5}{6}-\frac{2y}{6}\)
\(\Leftrightarrow\frac{3}{x}=\frac{5-2y}{6}\)
\(\Leftrightarrow x.\left(5-2y\right)=3.6\)
\(\Leftrightarrow x.\left(5-2y\right)=18\)
Mà \(x,y\in Z\Rightarrow5-2y\in Z\)
Lập bảng tìm nốt
\(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)
\(\Leftrightarrow\frac{2}{y}=\frac{x}{6}-\frac{1}{30}\)
\(\Leftrightarrow\frac{2}{y}=\frac{5x}{30}-\frac{1}{30}\)
\(\Leftrightarrow\frac{2}{y}=\frac{5x-1}{30}\)
\(\Leftrightarrow y(5x-1)=60\)
Làm nốt , đến đây dễ rồi
(x + 1) (y+2)=5
=> x = 0 và y = 3
(x+1)(y+2)=6
=> x = 0 và y = 4
(x+6) chia het cho x+2
a)\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)
=> \(\frac{2}{y}=\frac{x}{2}-\frac{1}{2}\)
=> \(\frac{2}{y}=\frac{x-1}{2}\)
=> \(y\left(x-1\right)=4\)
Vì x,y \(\inℕ\)nên x - 1 \(\inℕ\)=> y và x - 1 thuộc Ư(4)
Ta có : Ư(4) = {1;2;4}
Lập bảng :
y | 1 | 2 | 4 |
x - 1 | 4 | 2 | 1 |
x | 5 | 3 | 2 |
Vậy \(\left(x,y\right)\in\left\{\left(5,1\right);\left(3,2\right);\left(2,4\right)\right\}\)
b) \(\frac{5}{x}-\frac{y}{3}=\frac{1}{6}\)
=> \(\frac{5}{x}=\frac{1}{6}+\frac{y}{3}\)
=> \(\frac{5}{x}=\frac{1}{6}+\frac{2y}{6}\)
=> \(\frac{5}{x}=\frac{1+2y}{6}\)
=> \(x\left(1+2y\right)=30\)
Vì x,y thuộc N nên 1 + 2y thuộc N => x và 1 + 2y thuộc Ư(30)
Ta có : Ư(30) = {1;2;3;5;6;10;15;30}
Lập bảng :
x | 1 | 2 | 3 | 5 | 6 | 10 | 15 | 30 |
1 + 2y | 30 | 15 | 10 | 6 | 5 | 3 | 2 | 1 |
2y | 29 | 14 | 9 | 5 | 4 | 2 | 1 | 0 |
y | loại | 7 | loại | loại | 2 | 1 | loại | 0 |
Vậy : \(\left(x,y\right)\in\left\{\left(2,7\right);\left(6,2\right);\left(30,0\right)\right\}\)
c) Làm nốt
`6/y + 1/2=2`
`6/y = 2-1/2`
`6/y = 4/2 - 1/2`
`6/y = (4-1)/2`
`6/y = 3/2`
`=> y = 6 : 3/2`
`=> y = 4`
\(\dfrac{6}{y}+\dfrac{1}{2}=2\Leftrightarrow\dfrac{6}{y}=2-\dfrac{1}{2}\Leftrightarrow\dfrac{6}{y}=\dfrac{3}{2}\Leftrightarrow y=4\)