a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)

b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)

\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)

c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)

\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)

\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)

\(=\left(\dfrac{3}{7}\right)^{15}\)

\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)

\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)

a) 56:55+(49)0=56−5+1=5+1=656:55+(94​)0=56−5+1=5+1=6

b) (37)21:(1−4049)3(73​)21:(1−4940​)3

=(37)21:(949)3=(73​)21:(499​)3

=(37)21:[(37)2]3=(73​)21:[(73​)2]3

=(37)21:(37)6=(73​)21:(73​)6

=(37)21−6=(37)15=(73​)21−6=(73​)15

c) (23)3−(−523)0+49(

`@` `\text {Ans}`

`\downarrow`

`a)`

`-9/34*17/4`

`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)

`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)

`=`\(-\dfrac{9}{8}\)

`b)`

\(\dfrac{17}{15}\div\dfrac{4}{3}\)

`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)

`=`\(\dfrac{17}{20}\)

`c)`

\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)

`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)

a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)

\(=\dfrac{-9\cdot17}{34\cdot4}\)

\(=-\dfrac{153}{136}\)

\(=\dfrac{9}{8}\)

b) \(\dfrac{17}{15}:\dfrac{4}{3}\)

\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)

\(=\dfrac{17\cdot3}{15\cdot4}\)

\(=\dfrac{51}{60}=\dfrac{17}{20}\)

c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)

\(=\dfrac{21}{5}:-\dfrac{14}{5}\)

\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)

\(=\dfrac{21\cdot-5}{5\cdot14}\)

\(=-\dfrac{105}{70}=\dfrac{3}{2}\)

a) −934⋅17434−9​⋅417​

=−9⋅1734⋅4=34⋅4−9⋅17​

=−153136=−136153​

=98=89​

b) 1715:431517​:34​

=1715⋅34=1517​⋅43​

=17⋅315⋅4=15⋅417⋅3​

=5160=1720=6051​=2017​

c) 415:(−245)451​:(−254​)

=215:−145=521​:−514​

=215⋅−514=521​⋅−145​

=21⋅−55⋅14=5⋅1421⋅−5​

=−10570=32=−70105​=2

a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)

c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)

a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)

c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)

A = 2/3 + -1/3

    = 1/3

B = 25/11 . 13/12 . (-2,2)

    = 325/132 . (-2,2)

    = -65/12

C = 11/20 . -2/5

    = -11/50

Chúc bạn học tốt!! ^^

    = -

a, A = \(\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)\)

       =\(\dfrac{2}{3}+\dfrac{-1}{3}\)

       =\(\dfrac{1}{3}\)

b, B = \(2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)\)

       = \(\dfrac{23}{11}.\dfrac{13}{12}.\dfrac{-11}{5}\)

       = \(\dfrac{23.13.\left(-11\right)}{11.12.5}\)

       = \(\dfrac{-299}{60}\)

c, C =\(\left(\dfrac{3}{4}-0,2\right)\)\(.\left(0,4\right)-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\)\(.\dfrac{2}{5}-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)\(.\dfrac{-2}{5}\)

        = \(\dfrac{1}{4}.\dfrac{-2}{5}\)

        = \(\dfrac{-2}{20}=\dfrac{-1}{10}\)

Chúc bạn học tốt!

Bài giải:

a) 5³ : (-5)² = 5³ : 5² = 5^{3 – 2} = 5

b) ${\left(\frac{3}{4}\right)}^5$: ${\left(\frac{3}{4}\right)}^3$= ${\left(\frac{3}{4}\right)}^{5-3}$= ${\left(\frac{3}{4}\right)}^2$ = $\frac{9}{16}$

c) (-12)³ : 8³ = -12³ : 8³= (-4 . 3)³ : (4 . 2)³ = ${\left(\frac{-4.3}{4.2}\right)}^3$= ${\left(-\frac{3}{2}\right)}^3$= -$\frac{27}{8}$

a) 5³ : (-5)² = 5³ : 5² = 5^{3 – 2} = 5

b)( \(\dfrac{3}{4}\))⁵:(\(\dfrac{3}{4}\))³= \(\dfrac{3}{4}\)= )²= \(\dfrac{9}{16}\)

c) (-12)³ : 8³ = -12³ : 8³= (-4 . 3)³ : (4 . 2)³ = (\(\dfrac{-4\cdot3}{4\cdot2}\)= )³= \(\dfrac{-27}{8}\)

\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)

1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)

b:Ta có: \(\dfrac{1}{4}+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1-\dfrac{1}{4}\)

\(=-\dfrac{1}{8}-1=-\dfrac{9}{8}\)

\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)

a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)