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Ta có : \(A=\frac{4x+3}{x-2}=\frac{2\left(x-2\right)+7}{x-2}=2+\frac{7}{x-2}\)
Để \(A\in Z\)thì \(7⋮x-2\)hay x-2 là Ư(7)={1;-1;7;-7}
Do đó:
| x-2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy .....
Ta có : \(B=\frac{2x-15}{x+1}=\frac{2\left(x+1\right)-17}{x+1}=2-\frac{17}{x+1}\)
Để \(B\in Z\)thì \(17⋮x+1\)hay x+1 là Ư(17)={1;-1;17;-17}
Do đó :
| x+1 | 1 | -1 | 17 | -17 |
| x | 0 | -2 | 16 | -18 |
Vậy ................
\(\frac{1}{2}+\frac{5}{6}-\frac{3}{8}\)
\(=\frac{12}{24}+\frac{20}{24}-\frac{9}{24}\)
\(=\frac{23}{24}\)
\(\frac{10}{15}\cdot\frac{7}{4}+\frac{7}{4}\cdot\frac{9}{15}-\frac{4}{15}\cdot\frac{7}{4}\)
\(=\frac{7}{4}\cdot\left(\frac{10}{15}+\frac{9}{15}-\frac{4}{15}\right)\)
\(=\frac{7}{4}\cdot1=\frac{7}{4}\)
\(\left(\frac{4}{5}+\frac{1}{2}\right)\left(\frac{3}{13}-\frac{8}{13}\right)\)
\(=\left(\frac{8}{10}+\frac{5}{10}\right)\cdot\left(-\frac{5}{13}\right)\)
\(=\frac{13}{10}\cdot\left(-\frac{5}{13}\right)=-\frac{1}{2}\)
a, 3x-12 = 30
=> 3x = 30 + 12
=> 3x = 42
=> x = 42 : 3 = 14
Vậy x = 14
b, \(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}\div\frac{2}{3}\Rightarrow\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
c, 2x2 = 32
=> x2 = 32 : 2
=> x2 = 16
=> x2 = 42
=> x = 4
Vậy x = 4
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)