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\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\\ \Leftrightarrow6x-42=7y-42\\ \Leftrightarrow6x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{x-y}{7-6}=\dfrac{-4}{1}=-4\\ \dfrac{x}{7}=-4\Leftrightarrow x=-28\\ \dfrac{y}{6}=-4\Leftrightarrow y=-24\)
11: |2x-3|-1/3=0
=>|2x-3|=1/3
=>\(\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{10}{3}\\2x=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
12: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)
=>\(\left|x+\dfrac{1}{4}\right|=\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{12}\end{matrix}\right.\)
13: \(\left|x-1\right|-2x=\dfrac{1}{2}\)
=>\(\left|x-1\right|=2x+\dfrac{1}{2}\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}-x+1\right)\left(2x+\dfrac{1}{2}+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
14: \(3x-\left|x+15\right|=\dfrac{5}{4}\)
=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16.25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\)
=>\(x=8.125\)
Vì y = x + 1
<=> 3x + 5 . ( x + 1 ) = 13
<=> 3x + 5x + 5 = 13
<=> 8x = 13 - 5
<=> 8x = 8
<=> x = 8 : 8
<=> x = 1
Có x thì tìm y: y = x + 1
y = 1 + 1
y = 2
Vậy: x = 1, y = 2
Vì y = x + 1
<=> 3x + 5 . ( x + 1 ) = 13
<=> 3x + 5x + 5 = 13
<=> 8x = 13 - 5
<=> 8x = 8
<=> x = 8 : 8
<=> x = 1
Có x thì tìm y: y = x + 1
y = 1 + 1
y = 2
Vậy: x = 1, y = 2
\(xy+14+2y+7x=-10\)
\(\Rightarrow xy+7x+7y=-24\)
\(\Rightarrow x\left(y+7\right)+7y=-24\)
\(\Rightarrow x\left(y+7\right)+7y+49=-24+49\)
\(\Rightarrow x\left(y+7\right)+7\left(y+7\right)=25\)
\(\Rightarrow\left(x+7\right)\left(y+7\right)=25\)
\(\Rightarrow\left(x+7\right);\left(y+7\right)\inƯ\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
Xét bảng
x+7 | 1 | -1 | 5 | -5 | 25 | -25 |
y+7 | 25 | -25 | 5 | -5 | 1 | -1 |
x | 6 | -8 | -2 | -12 | 18 | -32 |
y | 18 | -32 | -2 | -12 | 6 | -8 |
Vậy.........................
\(xy+x+y=2\)
\(\Rightarrow x\left(y+1\right)+\left(y+1\right)=2+1\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=3\)
\(\Rightarrow\left(x+1\right);\left(y+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Xét bảng
x+1 | 1 | -1 | 3 | -3 |
y+1 | 3 | -3 | 1 | -1 |
x | 0 | -2 | 2 | -4 |
y | 2 | -4 | 0 | -2 |
Vậy.....................................
\(xy-10+5x-3y=2\)
\(\Rightarrow xy-5x-3y=12\)
\(\Rightarrow x\left(y-5\right)-3y+15=12+15\)
\(\Rightarrow x\left(y-5\right)-3\left(y-5\right)=27\)
\(\Rightarrow\left(x-3\right)\left(y-5\right)=27\)
\(\Rightarrow\left(x-3\right);\left(y-5\right)\inƯ\left(27\right)=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)
Tự xét bảng như trên
\(xy-1=3x+5y+4\)
\(\Rightarrow xy-3x-5y=4+1\)
\(\Rightarrow x\left(y-3\right)-5y+15=1+4+15\)
\(\Rightarrow x\left(y-3\right)-5\left(y-3\right)=20\)
\(\Rightarrow\left(x-5\right)\left(y-3\right)=20\)
\(\Rightarrow\left(x-5\right)\left(y-3\right)\inƯ\left(20\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
Xét bảng
x-5 | 1 | -1 | 2 | -2 | 4 | -4 | 5 | -5 | 10 | -10 | 20 | -20 |
y-3 | 20 | -20 | 10 | -10 | 5 | -5 | 4 | -4 | 2 | -2 | 1 | -1 |
x | 6 | 4 | 7 | 3 | 9 | 1 | 10 | 0 | 15 | -5 | 25 | -15 |
y | 23. | -17 | 13 | -7 | 8 | -2 | 7 | -1 | 5 | 1 | 4 | 2 |
Vậy......................................
3xy−2x+5y=−43xy−2x+5y=−4
⇒y(3x+5)−2x=−4⇒y(3x+5)−2x=−4
⇒y(3x+5)−2x+4=0⇒y(3x+5)−2x+4=0
⇒3x+5=0⇒3x+5=0 và 3y−2=03y−2=0
⇒x=−5/3⇒x=−5/3 và y=2/3