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`@` `\text {Ans}`
`\downarrow`
`1)`
`5/7*37 13/23 - 51 13/23*5/7`
`= 5/7* (37 13/23 - 51 13/23)`
`= 5/7* (-14)`
`= -10`
`2)`
`-2/3 +1/3+0,5+2 1/2`
`= -2/3 + 1/3 + 1/2 + 5/2`
`= (-2/3+1/3) + (1/2+5/2)`
`= -1/3 + 3`
`=8/3`
`3)`
`-0,5+2/3+1/2`
`= -1/2 + 2/3 + 1/2`
`= (-1/2 + 1/2) + 2/3`
`= 2/3`
`4)`
`(8+2 1/3-3/5) -(5+0,4)-(3 1/2 -2)`
`= 8+ 7/3 - 3/5 - 5 - 0,4 - 7/2 + 2`
`= (8+2-5) + (-3/5 - 2/5) + (7/3 - 7/2)`
`= 5 - 1 - 7/6`
`= 4 - 7/6 = 17/6`
`5)`
`(2/9-7/12):3/4+(16/9-5/12):3/4`
`= (2/9 - 7/12) \times 4/3 + (16/9 - 5/12) \times 4/3`
`= 4/3 *(2/9 - 7/12 + 16/9 - 5/12)`
`= 4/3 * [(2/9 + 16/9) + (-7/12 - 5/12)]`
`= 4/3 * ( 2 - 1)`
`= 4/3 * 1 = 4/3`
`6)`
`-(2021.0,7+19,75) +0,7- (8-19,75)`
`= -2021*0,7 -19,75 + 0,7 - 8 + 19,75`
`= 0,7*(-2021 + 1) - 8`
`= -1414-8`
`= -1422`
`7)`
`15/34+7/21+19/34-20/15`
`= (15/34 + 19/34) + 7/21 - 20/15`
`= 1 + 7/21 - 20/15`
`= 4/3 - 20/15 =0`
`8)`
`2 5/6+1/6:(-5/8)`
`= 17/6 + (-4/15)`
`= 77/30`
`9)`
`(-2)^2 +2/9. (4/5-2/3)`
`= 4 + 2/9*2/15`
`= 4+4/135`
`= 544/135`
`10)`
`(-1/5+3/7):5/4+(-4/5+4/7):5/4`
`= (-1/5+3/7) * 4/5 + (-4/5+4/7) * 4/5`
`= 4/5*(-1/5 +3/7-4/5+4/7)`
`= 4/5*[(-1/5-4/5)+(3/7+4/7)]`
`= 4/5* (-1+1)`
`= 4/5*0=0`
`11)`
`2022,2021 . 1954,1945+ 2022,2021 . (-1954,1945)`
`= 2022,2021 * [1954,1945 + (-1954,1945)]`
`= 2022,2021*0 `
`= 0`
`12)`
`-5,2 .72 +69,1 +5,2 . (-28)+(-1,1)`
`= -5,2*72 + 69,1 - 5,2*28 - 1,1`
`= -5,2*(72+28) + (69,1 - 1,1)`
`= -5,2*100 + 68`
`= -520 + 68`
`= -452`
`13)`
`(7 -1/2-3/4) : (5-1/4-5/8)`
`= 23/4 \div 33/8`
`=46/33`
`14)`
`(8+ 2 1/3 -3/5) -(5+0,4) -( 3 1/3 - 2)`
`= 8+ 2 1/3 - 3/5 - 5 - 0,4 - 3 1/3 + 2`
`= (8+2-5) + (2 1/3 - 3 1/3) - (0,6 + 0,4) `
`= 5 - 1 - 1`
`= 3`
a, ( 152 +và 2/4 - 148 và 3/8 ) : 0,2 = x : 0,3
=> 33/8 : 1/5 = x : 3/10
=> x : 3/10 = 165/8
=> x = 99/10
b, ( 85 và 7/30 - 83 và 5/18 ) : 2 và 2/3 = 0,01x : 4
=> 88/45 : 8/3 = 0,01x : 4
=> 0,01x : 4 = 11/15
=> 0,01x = 44/15
=> x = 880/3
c, x - 1/ x + 5 = 6/7
=> 7( x - 1 ) = 6( x + 5 )
=> 7x - 7 = 6x + 30
=> 7x - 6x = 7 + 30
=> x = 37
d, x2/6 = 24/25
=> x2. 25 = 6 . 24
=> x2.25 = 144
=> x2 = 144/25
=> x = ( 12/5)2 hoặc x = ( -12/5)
g, x - 3/ x + 5 = 5/7
=> 7( x - 3 ) = 5 ( x + 5 )
=> 7x - 21 = 5x + 25
=> 7x - 5x = 21 + 25
=> 2x = 46
=> x = 23
\(a,\frac{2^3.2^4}{2^5}=\frac{2^7}{2^5}=2^2=4\)
\(b,\frac{\left(0,2\right)^5.\left(0,6\right)^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{\left(0,2\right)^5.\left(0,3\right)^4.2^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{2^4}{\left(0,2\right)^2}\)
\(c,\frac{3^3.12^4}{6^5.9^4}=\frac{3^3.6^4.2^4}{6^5.3^8}=\frac{2^4}{6.3^5}=\frac{2^4}{2.3.3^5}=\frac{2^3}{3^6}\)
\(d,\frac{2^3+2^4+2^5}{7^2}=\frac{8+16+32}{49}=\frac{56}{49}=\frac{8}{7}\)
\(e,\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
A = 6 -2/3 + 1 /2 - 5 -5/3 +3/2 -3 + 7/3 - 5/2
= (6 - 5 - 3) - ( 2/3 -5/3 + 7/3 ) + ( 1/2 +3/2 - 5/2)
= -2 + 0 -1/2 = -5/2
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
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