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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
a)ta có xy=7*9=7*3*3
vậy x =9;21 , y=7;3
b) xy=-2*5
mà x<0<y
nên x=-2 ,y=5
c)x-y=5 hay x=y+5
\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)
a)Ta có: \(\frac{x}{18}=\frac{3}{y}\)
\(\Rightarrow x\times y=18\times3=54\)mà \(x< y< 0\),\(x\in Z\)
\(\Rightarrow x\times y=-54\times-1=-27\times-2=-18\times-3=-9\times-6\)
Vậy (x;y)\(\in\){(-54;-1),(-27;-2),(-18;-3),(-9;-6)}
Câu b) tương tự vậy nha!
a) ta có : \(x-y=7\Leftrightarrow x=7+y\)
thay vào \(x+y=3\Leftrightarrow7+y+y=3\Leftrightarrow y+y=3-7=-4\)
\(\Leftrightarrow2y=-4\Leftrightarrow y=\dfrac{-4}{2}=-2\)
ta có : \(y=-2\Rightarrow x=7+y=7-2=5\) vậy \(y=-2;x=5\)
b) ta có : \(3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3=1\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=3\\y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=-1\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=-3\\y-2=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y-1\end{matrix}\right.\end{matrix}\right.\) (tmđk)
vậy ........................................................................................