rút gọn -115.137/115.138
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\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}.\left(-3\right)^2\)
\(=\dfrac{2}{3}+\dfrac{5}{6}.\dfrac{1}{5}-\dfrac{1}{18}.9\)
\(=\dfrac{2}{3}+\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{4}{6}+\dfrac{1}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}\cdot\left(-3\right)^2\)
\(=\dfrac{4}{6}+\dfrac{5}{6}:5-\dfrac{1}{18}\times9\)
\(=\dfrac{4}{6}+\dfrac{5}{6}\times\dfrac{1}{5}-\dfrac{1}{18}\times\dfrac{9}{1}\)
\(=\dfrac{4}{6}+\dfrac{5}{30}-\dfrac{9}{18}\)
\(=\dfrac{20}{30}+\dfrac{5}{30}-\dfrac{1}{2}\)
\(=\dfrac{25}{30}-\dfrac{1}{2}\)
\(=\dfrac{25}{30}-\dfrac{15}{30}\)
\(\dfrac{10}{30}=\dfrac{1}{3}\)
\(\dfrac{2}{9}\) - \(\dfrac{7}{8}\).\(x\) = \(\dfrac{1}{3}\)
\(\dfrac{7}{8}\).\(x\) = \(\dfrac{2}{9}\) - \(\dfrac{1}{3}\)
\(\dfrac{7}{8}\).\(x\) = - \(\dfrac{1}{9}\)
\(x\) = - \(\dfrac{1}{9}\): \(\dfrac{7}{8}\)
\(x\) = - \(\dfrac{8}{63}\)
\(\dfrac{2}{9}-\dfrac{7}{8}\cdot x=\dfrac{1}{3}\)
\(\dfrac{7}{8}\cdot x=\dfrac{2}{9}-\dfrac{1}{3}\)
\(\dfrac{7}{8}\cdot x=\dfrac{2}{9}-\dfrac{3}{9}\)
\(\dfrac{7}{8}\cdot x=\dfrac{2-3}{9}\)
\(\dfrac{7}{8}\cdot x=\dfrac{-1}{9}\)
\(x=\dfrac{-1}{9}:\dfrac{7}{8}\)
\(x=\dfrac{-1}{9}\times\dfrac{8}{7}\)
\(x=\dfrac{-8}{63}\)
Lời giải:
$\frac{4}{x}+\frac{x}{3}=\frac{5}{6}$
$\frac{12+x^2}{3x}=\frac{5}{6}$
$12+x^2=\frac{5}{2}x$
$2x^2-5x+24=0$
$x^2+(x-2,5)^2=-17,75<0$ (vô lý)
Do đó không tồn tại $x$ thỏa mãn.
\(P=2.5+5.8+8.11+...+101.104\)
\(\Rightarrow9P=2.5.9+5.8.9+8.11.9+...+101.104.9\)
\(\Rightarrow9P=2.5.9+5.8.\left(11-2\right)+8.11.\left(14-5\right)+...+101.104.\left(107-98\right)\)
\(\Rightarrow9P=2.5.9-2.5.8+5.8.11-5.8.11+8.11.14+...-98.101.104+101.104.107\)
\(\Rightarrow9P=2.5.9-2.5.8+101.104.107\)
\(\Rightarrow9P=1123938\)
\(\Rightarrow P=124882\)
Sửa đề: \(\dfrac{2}{17}+\dfrac{15}{23}+\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)
\(=\left(\dfrac{2}{17}+\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)
\(=1+1+\dfrac{4}{19}\)
\(=2+\dfrac{4}{19}\)
\(=\dfrac{42}{19}\)
\(#Tmiamm\)
a; \(\dfrac{6}{x}\) < \(\dfrac{x}{7}\) < \(\dfrac{8}{x}\)
vì \(x\) \(\in\) N* ta có: 6.7 < \(x.x\) < 7.8
42 < \(x^2\) < 56
\(x^2\) = 49
\(x\) = \(\pm\) 7
Vì \(x\) \(\in\) N*; \(x\) = 7
b; \(\dfrac{x}{11}\) < \(\dfrac{12}{x}\) < \(\dfrac{x}{9}\)
9.12< \(x^2\) < 11.12
108 < \(x^2\) < 132
\(x^2\) = 121
\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)
Vì \(x\in\) N*
\(x\) = 11
S = 63 + 65 + 67 + ... + 6101
62 x S = 65 + 67 + 69 + ... + 6103
36S-S = 65 + 67 + 69 + ... + 6103 - 63 - 65 - 67 - ... - 6101
35S = 6103 - 63
S = ( 6103 - 63 ) : 35
CHÚC BẠN HỌC TỐT NHÉ!!
\(\dfrac{-11^5.13^7}{11^5.13^8}=\dfrac{-1.11^5.13^7}{11^5.13^8}=-\dfrac{1}{13}\)
\(\dfrac{-11^5\cdot13^8}{11^5\cdot13^7}\)
\(=\dfrac{-11^5\cdot13^7\cdot13}{11^5\cdot13^7}\)
\(=\dfrac{-11^5\cdot13}{11^5}\)
\(=\dfrac{11^5\cdot\left(-13\right)}{11^5}\)
\(=-13\)