(x+1) (y 2- 5) =6
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Khách
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11 tháng 1 2023
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>x+1=1 và y-2=1/2
=>x=0 và y=5/2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)
=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6
=>x-2y=9 và 2x-y=12
=>x=5; y=-2
c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
3 tháng 7 2021
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!
RG
0
28 tháng 8 2020
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 )
HN
30 tháng 5 2021
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 :)
CG
1
NM
Nguyễn Minh Quang
Giáo viên
12 tháng 8 2021
a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)
\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)
\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)
b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
30 tháng 3 2016
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
12 tháng 7 2023
2:
a: 5/x-y/3=1/6
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)
=>30-2xy=x
=>x(2y+1)=30
=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}
=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}
b: x/6-2/y=1/30
=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)
=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)
=>5xy-60=y
=>y(5x-1)=60
=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)
=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}