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\(\left(3x-15\right)^7=0\)
=> 3x - 15 = 0
=> 3x = 0 + 15
=> 3x = 15
=> x = 15 : 3
=> x = 5
vậy___
2xy +3x =2y +9
<=>2xy - 2 + 3x -3 = 9-3
<=>2y(x-1) +3(x-1) = 6
<=>(x-1)(2y+3) =6
ta có bảng:
| x-1 | 2 | 3 | -2 | -3 | 1 | 6 | -1 | -6 |
| 2y+3 | 3 | 2 | -3 | -2 | 6 | 1 | -6 | -1 |
| x | 3 | 4 | -1 | -2 | 2 | 7 | 0 | -5 |
| y | 0 | -1/2 | -3 | -5/2 | 3/2 | -1 | -9/2 | -2 |
Vậy: \(\left(x,y\right)\in\) {(3,0);(-1,-3)...bạn tự liệt kê ra}
17 + ( 26 - 3 x ) : 5 = 75
17 + ( 26 - 3x = 75 : 5
17 + ( 26 - 3x ) = 15
26 - 3x = 17 - 15
26 - 3x = 2
3x =26 -2
3x = 24
x = 24 : 3
x = 8
[3x-1]5=[3x-1]8
=>[3x-1]8-[3x-1]5=0
[3x-1]5.[[3x-1]3-1]=0
=>[3x-1]5 =0 hoặc [3x-1]3-1=0
=>3x-1=0 hoặc3x-1 =1
=>x=1/3 hoac x=2/3
Vay....
=>
\(xy+14+2y+7x=-10\)
\(\Rightarrow xy+7x+7y=-24\)
\(\Rightarrow x\left(y+7\right)+7y=-24\)
\(\Rightarrow x\left(y+7\right)+7y+49=-24+49\)
\(\Rightarrow x\left(y+7\right)+7\left(y+7\right)=25\)
\(\Rightarrow\left(x+7\right)\left(y+7\right)=25\)
\(\Rightarrow\left(x+7\right);\left(y+7\right)\inƯ\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
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......................................
=(2/7+4/9)+((-15/23)+(-8/23))+15/7
=(18/63+28/63)+(-1)+15/7
=46/63+(-1)+15/7
=46/63+(-63/63)+135/63
=118/63