Tính
a,\(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
b,\(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)
c, \(3\frac{1}{2}.\frac{4}{49}-\left[2,\left(4\right).2\frac{5}{11}\right]:\frac{-42}{53}\)
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TH1 :
\(\hept{\begin{cases}-x+2>0\\x-10< 0\end{cases}}\)
\(\hept{\begin{cases}-x>-2\\x< 10\end{cases}}\)
\(\hept{\begin{cases}x< 2\\x< 10\end{cases}\Rightarrow x< 2}\)
TH2 :
\(\hept{\begin{cases}-x+2< 0\\x-10>0\end{cases}}\)
\(\hept{\begin{cases}-x< -2\\x>10\end{cases}}\)
\(\hept{\begin{cases}x>2\\x>10\end{cases}\Rightarrow x>10}\)
\(\frac{13}{42}-\left(\frac{55}{42}-78\right)\left(-\frac{1}{2}\right)^3\)
\(=\frac{13}{42}-\left(\frac{55}{42}-78\right)\left(-\frac{1}{8}\right)\)
\(=\frac{13}{42}+\frac{3221}{42}.\left(-\frac{1}{8}\right)\)
\(=\frac{13}{42}+-\frac{3221}{42}\)
\(=-\frac{1604}{21}\)
Ta có: \(\left(a+2b-3c-d\right)\left(a+2b+3c+d\right)\)
\(=\left[\left(a+2b\right)-\left(3c+d\right)\right]\cdot\left[\left(a+2b\right)+\left(3c+d\right)\right]\)
\(=\left(a+2b\right)^2-\left(3c+d\right)^2\)
\(=a^2+4ab+4b^2-9c^2-6cd-d^2\)
( a + 2b - 3c - d )( a + 2b + 3c + d )
= [ ( a + 2b ) - ( 3c + d ) ][ ( a + 2b ) + ( 3c + d ) ]
= ( a + 2b )2 - ( 3c + d )2
= a2 + 4ab + 4b2 - ( 9c2 + 6cd + d2 )
= a2 + 4ab + 4b2 - 9c2 - 6cd - d2
172x2 - 79 : 983 = 2-3
<=> 172x2 - 79 : ( 2 . 72 )3 = 1/8
<=> 172x2 - 79 : ( 23 . 76 ) = 1/8
<=> 172x2 - 79 : 23 : 76 = 1/8
<=> 172x2 - 73 : 23 = 1/8
<=> 172x2 - ( 7 : 2 )3 = 1/8
<=> 172x2 - 343/8 = 1/8
<=> 172x2 = 43
<=> x2 = 43/172 = 1/4
<=> x = ±1/2
a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)
\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)
b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)
\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)
\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)
c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)
\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)
\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)