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a) (x+2)(y-3)=0
\(\orbr{\begin{cases}x+2=0\\y-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}}\)
a/ (x + 2) . ( y - 3 ) = 0
Nên: x + 2 = 0 => x = 0 - 2 = - 2
Hoặc y - 3 = 0 => y = 0+ 3 = 3
Vậy x = -2 hoặc y = 3
b/ (x + 4) . ( y - 2 ) = 2
Nên:
(-) x + 4 = 1 => ..........
y - 2 = 2 => ....
(-) x + 4 = 2 => ....
y - 2 = 1 => ....
(-) x + 4 = -1 =>....
y - 2 = -2 => ....
(-) x + 4 = -2
y - 2 = -1 => .......
* Các trường hợp kia tương tự
x+y=-10
=>x=-10-y
xy=24
\(\Leftrightarrow y\left(-10-y\right)=24\)
\(\Leftrightarrow y^2+10y+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-4\\y=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-4\end{matrix}\right.\)
\(\frac{3+x}{7+y}=\frac{3}{7};x+y=20\)
\(\Leftrightarrow21+7x=21+3y\Leftrightarrow7x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c dãy tỉ số ''='' nhau ta có
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{2}{10}=\frac{1}{5}\)
\(\Leftrightarrow\frac{x}{3}=\frac{1}{5}\Leftrightarrow5x=3\Leftrightarrow x=\frac{3}{5}\)
\(\Leftrightarrow\frac{y}{7}=\frac{1}{5}\Leftrightarrow5y=7\Leftrightarrow y=\frac{7}{5}\)
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 :)
1a) \(\frac{x-3}{x+7}=\frac{-5}{-6}\)
=> \(\frac{x-3}{x+7}=\frac{5}{6}\)
=> (x - 3).6 = 5.(x + 7)
=> 6x - 18 = 5x + 35
=> 6x - 5x = 35 + 18
=> x = 53
b) \(\frac{x-7}{x+3}=\frac{4}{3}\)
=> (x - 7). 3 = (x + 3). 4
=> 3x - 21 = 4x + 12
=> 3x - 4x = 12 + 21
=> -x = 33
=> x = -33
c) \(\frac{x-10}{6}=-\frac{5}{18}\)
=> (x - 10) . 18 = -5 . 6
=> 18x - 180 = -30
=> 18x = -30 + 180
=> 18x = 150
=> x = 150 : 18 = 25/3
d) \(\frac{x-2}{4}=\frac{25}{x-2}\)
=> (x - 2)(x - 2) = 25 . 4
=> (x - 2)2 = 100
=> (x - 2)2 = 102
=> \(\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=-8\end{cases}}\)
e) \(\frac{7}{x}=\frac{x}{28}\)
=> 7 . 28 = x . x
=> 196 = x2
=> x2 = 142
=> \(\orbr{\begin{cases}x=14\\x=-14\end{cases}}\)
f) \(\frac{40+x}{77-x}=\frac{6}{7}\)
=> (40 + x) . 7 = (77 - x).6
=> 280 + 7x = 462 - 6x
=> 280 - 462 = -6x + 7x
=> -182 = x
=> x = -182
( x- 7 ) . ( y + 3 ) = ?
Đề thiếu rồi bạn ơi
thank nha mk viết thiếu