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\(xy+2x+y-13=0\)
\(\Leftrightarrow x\left(y+2\right)+\left(y+2\right)=15\)
\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=15\)
TH1 : \(\left[{}\begin{matrix}x+1=1\\y+2=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=13\end{matrix}\right.\)
TH2 : \(\left[{}\begin{matrix}x+1=-1\\y+2=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\y=-17\end{matrix}\right.\)
TH3 : \(\left[{}\begin{matrix}x+1=15\\y+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=14\\y=-1\end{matrix}\right.\)
TH4 : \(\left[{}\begin{matrix}x+1=-15\\y+2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-16\\y=-3\end{matrix}\right.\)
Vậy .........................
\(35-\left[\left(2x-3\right)^2:7\right]=28\)
\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)
\(\Rightarrow\left(2x-3\right)^2:7=7\)
\(\Rightarrow\left(2x-3\right)^2=1\)
\(\Rightarrow2x-3=\pm1\)
\(\Rightarrow x=2\) hay \(x=1\)
35 - [(2\(x\) - 3)2:7 ] = 28
(2\(x-3\))2 : 7 = 35 - 28
(2\(x\) - 3)2 : 7 = 7
(2\(x\) - 3)2 = 7 \(\times\) 7
(2\(x-3\))2 = 72
\(\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-7+3\\2x=7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy \(x\in\) {-2; 5}
a) \(\left|2x-4\right|+\left|x-2y\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|2x-4\right|=0\\\left|x-2y\right|=0\end{matrix}\right.\)
+) \(\left|2x-4\right|=0\Rightarrow2x-4=0\Rightarrow2x=4\Rightarrow x=2\)
+) \(\left|x-2y\right|=0\Rightarrow x-2y=0\Rightarrow x=2y\Rightarrow2y=2\Rightarrow y=1\)
Vậy \(x=2;y=1\)
b) \(\left(x-1\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[\begin{matrix}\left(x-1\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
+) \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(x-y\right)^2=0\Rightarrow x-y=0\Rightarrow x=y=1\)
Vậy x = y = 1
\(3x+2x=35\)
\(\Rightarrow x\cdot\left(3+2\right)=35\)
\(\Rightarrow x\cdot5=35\)
\(\Rightarrow x=35:5\)
\(\Rightarrow x=7\)
Vậy: x = 7
a) \(xy+2x+3y=-6\)
\(\Rightarrow x\left(y+2\right)+3y+6=0\)
\(\Rightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(x=-3;y=-2\)
\(xy+2x+3y=-6\)
\(\Leftrightarrow xy+2x+3y+6=0\)
\(\Leftrightarrow y\left(x+3\right)+\text{2}\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Leftrightarrow\left\{\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
2x - 7m + y = (2x + y) - 7m
Ta có: m = 2 => 7m = 7.2 = 14
Ta có: (2x + y) - 14
Thế 2x + y = 35 vào (2x + y) - 14 (hay 7m)
Ta được: 35 - 14 = 21
Vậy 2x - 7m + y = 21 khi 2x + y = 35 và m = 2
\(2x-7m+y=\left(2x+y\right)-7m=35-7.2=35-14=21\)