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\(\frac{13}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)< x< \frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow\frac{13}{6}.\frac{-1}{3}< x< \frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{18}< x< \frac{-11}{18}\)
=> Ko có số nguyên x nào thỏa mãn điều kiện trên
Câu 1:
a) = \(\dfrac{-7}{2}\) x \(\dfrac{45}{32}\) = \(\dfrac{-315}{64}\)
b) = \(\dfrac{18}{7}\) : \(\dfrac{-27}{14}\) = \(\dfrac{18}{7}\) x \(\dfrac{14}{-27}\) = \(\dfrac{-4}{3}\)
c) = \(\dfrac{-3}{8}\) x ( \(\dfrac{5}{11}\) + \(\dfrac{6}{11}\) + 2 ) = \(\dfrac{-3}{8}\) x 3 = \(\dfrac{-9}{8}\)
Câu 2:
\(\dfrac{-3}{5}\) . x + \(\dfrac{7}{6}\) = \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{5}{4}\) - \(\dfrac{7}{6}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{1}{12}\)
\(\Leftrightarrow\) x = \(\dfrac{1}{12}\) : \(\dfrac{-3}{5}\)
\(\Leftrightarrow\) x = \(\dfrac{-5}{36}\)
Với \(x\le-5\Rightarrow\left\{{}\begin{matrix}VT>0\\VP\le0\end{matrix}\right.\) BPT vô nghiệm
Với \(x>-5\) hai vế dương, bình phương:
\(x^2+x+4< \left(x+5\right)^2\)
\(\Leftrightarrow x^2+x+4< x^2+10x+25\)
\(\Rightarrow9x>-21\Rightarrow x>-\frac{7}{3}\)
Mà x nguyên nên \(-2\le x\le10\)
Có \(10-\left(-2\right)+1=13\) giá trị thỏa mãn
a: \(=\dfrac{-3}{7}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+2+\dfrac{3}{7}=2\)
b: \(=-\dfrac{5}{7}:\left(24-\dfrac{166}{7}\right)+\dfrac{37}{3}\)
\(=-\dfrac{5}{7}:\dfrac{2}{7}+\dfrac{37}{3}=\dfrac{-5}{2}+\dfrac{37}{3}=\dfrac{59}{6}\)
c: \(=4-\dfrac{32}{27}\cdot\dfrac{-27}{8}=4+4=8\)
d: \(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\dfrac{11+5}{20}\cdot\dfrac{5}{7}\)
\(=\dfrac{7}{5}-\dfrac{6}{20}\cdot\dfrac{5}{7}=\dfrac{29}{35}\)
1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)
\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)
\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)
Bài 2:
a: \(=248+2064-12-236\)
\(=12-12+2064=2064\)
b: \(=-298-302-300=-600-300=-900\)
c: \(=5-7+9-11+13-15=-2-2-2=-6\)
d: \(=456+58-456-38=20\)
+) 4M = \(\dfrac{4^{13}+4}{4^{13}+1}=1+\dfrac{3}{4^{13}+1}\)
+) 4N = \(\dfrac{4^{14}+4}{4^{14}+1}=1+\dfrac{3}{4^{14}+1}\)
Có 413+1 < 414+1
⇒ \(\dfrac{3}{4^{13}+1}\) > \(\dfrac{3}{4^{14}+1}\)
⇒ \(1+\dfrac{3}{4^{13}+1}\) > \(1+\dfrac{3}{4^{14}+1}\)
⇒ 4M > 4N
⇒ M > N
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