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Giải:
a) \(\left(x-4\right).\left(y+1\right)=8\)
\(\Rightarrow\left(x-4\right)\) và \(\left(y+1\right)\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Ta có bảng giá trị:
| x-4 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| y+1 | -1 | -2 | -4 | -8 | 8 | 4 | 2 | 1 |
| x | -4 | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| y | -2 | -3 | -5 | -9 | 7 | 3 | 1 | 0 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
Vậy \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
b) \(\left(2x+3\right).\left(y-2\right)=15\)
\(\Rightarrow\left(2x+3\right)\) và \(\left(y-2\right)\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
| 2x+3 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| y-2 | -1 | -3 | -5 | -15 | 15 | 5 | 3 | 1 |
| x | -9 | -4 | -3 | -2 | -1 | 0 | 1 | 6 |
| y | 1 | -1 | -3 | -13 | 17 | 7 | 5 | 3 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
Vậy \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
c) \(xy+2x+y=12\)
\(\Rightarrow x.\left(y+2\right)+\left(y+2\right)=14\)
\(\Rightarrow\left(x+1\right).\left(y+2\right)=14\)
\(\Rightarrow\left(x+1\right)\) và \(\left(y+2\right)\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)
| x+1 | 1 | 2 | 7 | 14 |
| y+2 | 14 | 7 | 2 | 1 |
| x | 0 | 1 | 6 | 13 |
| y | 12 | 5 | 0 | -1 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\)
Vậy \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\)
d) \(xy-x-3y=4\)
\(\Rightarrow y.\left(x-3\right)-\left(x-3\right)=7\)
\(\Rightarrow\left(y-1\right).\left(x-3\right)=7\)
\(\Rightarrow\left(y-1\right)\) và \(\left(x-3\right)\inƯ\left(7\right)=\left\{1;7\right\}\)
Ta có bảng giá trị:
| x-3 | 1 | 7 |
| y-1 | 7 | 1 |
| x | 4 | 10 |
| y | 8 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(4;8\right);\left(10;2\right)\right\}\)
\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)
\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)
\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)
\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)
\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)
\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)
\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\) ( 4^3 = 8^2 ; 5^4 = 25^2 )
\(=\frac{1}{25\times7}=\frac{1}{175}\)
\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)
\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)
\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)
\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)
Ta có :
a, \(\frac{31}{12}-(\frac{2}{5}+x)=\frac{2}{3}\)
\(\Rightarrow\frac{2}{5}+x=\frac{31}{12}-\frac{2}{3}=\frac{23}{12}\)
\(\Rightarrow\frac{23}{12}-\frac{31}{12}=\frac{-8}{12}=\frac{-2}{3}\)
Câu b để mk làm sau
\(x+\frac{2}{15}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{2}{15}\)
\(x=\frac{1}{5}\)
h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)
\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}:x=\frac{1}{12}\)
\(x=\frac{2}{3}:\frac{1}{12}\)
\(x=8\)
1)\(\left(2^5:2^3\right).2^x=64\)
\(\Rightarrow2^{5-3+x}=2^6\)
\(\Rightarrow2^{2+x}=2^6\)
\(\Rightarrow.2^22^x=2^6\)
\(\Rightarrow2^x=2^6:2^2\)
\(\Rightarrow2^x=2^4\Rightarrow x=4\)
2)Tính:
\(F=3^0+3^1+...+3^9\)
\(\Rightarrow3F=3\left(3^0+3^1+...+3^9\right)=3+3^2+3^3+...+3^{10}\)
\(3F-F=3+3^2+...+3^{10}-3^0-3^1-...-3^9\)
\(2F=3^{10}-3^0=3^{10}-1\)
\(F=\frac{3^{10}-1}{2}\)
2
ta có : F = 1 + 3 + 32 + ..... + 39
=> 3F = 3 + 32 + 33 +..... + 310
=> 3F - F = 310 - 1
=> 2F = 310 - 1
=> F = \(\frac{3^{10}-1}{2}\)
a) \(-39-\left(15+x\right)=18\)
\(-39-15-x=18\)
\(-54-x=18\)
\(x=-54-18\)
\(x=-72\)
vậy \(x=-72\)
b) \(31-\left(13+x\right)=18.\left(-4\right)\)
\(31-13-x=-72\)
\(18-x=-72\)
\(x=18+72\)
\(x=90\)
vậy \(x=90\)
c) \(58-\left(47+3x\right)=-18-2x\)
\(58-47-3x+2x=-18\)
\(11-x=-18\)
\(x=11+18\)
\(x=29\)
vậy \(x=29\)
d) \(37-\left(115+x\right)=188\)
\(37-115-x=188\)
\(-78-x=188\)
\(x=-78-188\)
\(x=-266\)
vậy \(x=-266\)
a)
\(-39-\left(15+x\right)=18\)
\(\Rightarrow15+x=-39-18\)
\(\Rightarrow15+x=-57\)
\(\Rightarrow x=-57-15\)
\(\Rightarrow x=-72\)
Vậy \(x=-72\)
b)
\(31-\left(13+x\right)=18.\left(-4\right)\)
\(\Rightarrow31-\left(13+x\right)=-72\)
\(\Rightarrow13+x=31-\left(-72\right)\)
\(\Rightarrow13+x=103\)
\(\Rightarrow x=103-13\)
\(\Rightarrow x=90\)
Vậy \(x=90\)
Tìm x
a, 2x - (3-x) = -18
\(\Leftrightarrow2x-3+x=-18\)
\(\Leftrightarrow2z+x=-18+3\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=15:3=5\)
ok