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7,5x . 519 = 5 20 . 511
7,5x . 519 = 531
7,5x =531:519
7,5x=512
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A = 1 x 2 + 2 x 3 + 3 x 4 + ... + 19 x 20
=> 3A = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ... + 19 x 20 x ( 21 - 18 )
=> 3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 19 x 20 x 21 - 18 x 19 x 20
=> 3A = 19 x 20 x 21
=> A = 19 x 20 x 7
Vậy A = 19 x 20 x 7
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\(2,\frac{9}{x}=\frac{2}{5}-\frac{7}{20}\)
\(\Rightarrow\frac{9}{x}=\frac{1}{20}\)
\(\Rightarrow x=9.20\)
\(\Rightarrow x=180\)
\(\frac{x}{5}=\frac{5}{6}+\left(-\frac{19}{30}\right)\)
\(\frac{\Rightarrow x}{5}=\frac{1}{5}\)
\(\Rightarrow x=1\)
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2)
\(2+\dfrac{5}{7}+\left(\dfrac{\dfrac{3}{19}+\dfrac{3}{23}-\dfrac{3}{28}}{\dfrac{5}{19}+\dfrac{5}{23}-\dfrac{5}{28}}\right)\cdot x=\dfrac{20}{7}\\ \left[\dfrac{3\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}{5\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}\right]\cdot x=\dfrac{20}{7}-\dfrac{5}{7}-2\\ \dfrac{3}{5}x=\dfrac{15}{7}-2\\ \dfrac{3}{5}x=\dfrac{1}{7}\\ x=\dfrac{5}{21}\)
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Tìm x :
5x . 519 = 520 . 511
520 . 511 = 520 + 11
520 . 511 = 531
5x = 531 : 519
5x = 531 - 19
5x = 512
x = 12
Vậy : x = 12
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Ta có: \(A=\frac {19^{20}+5}{19^{20}-8}=\frac {19^{20}-8+13}{19^{20}-8}=1+\frac {13}{19^{20}-8}\)
\(B=\frac {19^{20}+6}{19^{20}-7}=\frac {19^{20}-7+13}{19^{20}-7}=1+\frac {13}{19^{20}-7}\)
Vì \(19^{20}-8<19^{20}-7\) nên \(\frac {13}{19^{20}-8}>\frac {13}{19^{20}-7}\)
\(\Rightarrow\)\(1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\) Hay \(A>B\)
Vậy A>B
ta có A = \(\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
và B = \(\frac{19^{20}+6}{19^{20}-7}=\frac{19^{20}-7+13}{19^{20}-7}=1+\frac{13}{19^{20}-7}\)
vì \(\frac{13}{19^{20}-8}>\frac{13}{19^{20}-7}\)\(\Rightarrow1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\)\(\Rightarrow A>B\)
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|x - 5| + 19 = 20
|x -5 | = 20 - 19
| x - 5| = 1
\(\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)
\(x\in\) { 4; 6}
\(|x-5|=1\)
\(\left\{{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)