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a: \(A=\dfrac{1.3-26}{2.6}-\dfrac{3}{8}\cdot\dfrac{1}{2}=-\dfrac{19}{2}-\dfrac{3}{16}=-\dfrac{155}{16}\)
b: \(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{1}{3}\cdot\dfrac{13}{4}=\dfrac{13}{12}\)
c: Để A<x<B thì \(-\dfrac{155}{16}< x< \dfrac{13}{12}\)
=>-10<x<2
hay \(x\in\left\{-9;-8;-7;...;0;1\right\}\)
\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)
\(=-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)
\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)
Ta có : \(A=-\frac{359}{390}\approx-0,9\)
\(B=\frac{13}{12}\approx1,08\)
\(\Rightarrow A< x< B\) mà x nguyên \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có:
\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{\frac{-13}{10}}{\frac{13}{5}}-\frac{5}{6}:2=\frac{-1}{2}-\frac{5}{12}=\frac{-11}{12}\)
\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}:\frac{75}{26}=\frac{13}{12}\)
\(\Rightarrow A< x< B\Rightarrow\frac{-11}{12}< x< \frac{13}{12}\Rightarrow-1< x\le1\Rightarrow x\in\left\{0;1\right\}\)
a) \(\left(\frac{16}{2}\right)^x=2\)
\(8^x=2\)
<=> 23x = 21
<=> 3x = 1
=> x = \(\frac{1}{3}\)
b) 8x : 2x = 4
(8 : 2)x = 4
4x = 4
=> x = 1
c) \(\left(\frac{1}{2}\right)^x=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
=> x = 5
A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)
=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)
B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)
b) ta có : A=\(-\frac{359}{390}\approx-0,9\)
B= \(\frac{13}{12}\approx1,08\)
=> A<x<B mà x nguyên => x=0 hoặc x=1
\(B=\left(-13\cdot\frac{2}{5}+\frac{-2}{9}\div2\frac{1}{2}+\frac{2}{5}\cdot\frac{11}{9}\right)\cdot2\frac{1}{2}\)
\(B=\left(13\cdot\frac{2}{5}+\frac{-2}{9}\cdot\frac{2}{5}+\frac{2}{5}\cdot\frac{11}{9}\right)\cdot\frac{5}{2}\)
\(B=\left[\frac{2}{5}\cdot\left(-13+\frac{-2}{9}+\frac{11}{9}\right)\right]\cdot\frac{5}{2}\)
\(B=\left[\frac{2}{5}\cdot\left(-12\right)\right]\cdot\frac{5}{2}\)
\(B=\frac{-24}{5}\cdot\frac{5}{2}=-12\)
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