Ta có:-0,7(43⁴³-17¹⁷)=[-7(43⁴³-17¹⁷)]:10

*43⁴³=43³.43⁴⁰=43³.(43⁴)¹⁰=(...7).(...1)¹⁰=(...7).(...1)=...7 (1)

*17¹⁷=17¹⁶.17=(17⁴)⁴.17=(...1)⁴.17=(....1).17=....7 (2)

Từ (1) và (2) suy ra 43⁴³-17¹⁷=(...7)-(...7)=.....0 chia hết cho 10

Vậy -0,7(43⁴³-17¹⁷) là 1 số nguyên

\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)

\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)

\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)

\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)

\(A=1-1=0\)

\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)

\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)

\(A=0-\frac{43}{101}=\frac{-43}{101}\)

\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)

\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)

\(C=\frac{-2}{1}=-2\)

Ta có:

\(\frac{-17}{35}=\frac{-289}{595}.\)

\(\frac{-43}{85}=\frac{-301}{595}.\)

Vì \(-289>-301\)

\(\Rightarrow\frac{-289}{595}>\frac{-301}{595}.\)

\(\Rightarrow\frac{-17}{35}>\frac{-43}{85}.\)

Chúc bạn học tốt!

Ta có 43\(^1\) = 43

43\(^2\) = \(\overline{.......9}\) (tận cùng là 9)

43\(^3\) = \(\overline{........7}\);

43\(^4\) = \(\overline{........1}\);

43\(^3\) = \(\overline{........3}\)

=>43\(^{4k}\) =\(\overline{........1}\)

43\(^{4k+1}\) = \(\overline{........3}\)

43\(^{4k+2}\)= \(\overline{.......9}\)

43\(^{4k+3}\) = \(\overline{........7}\)

Mà 43 = 4.10 + 3 => 43\(^{43}\) = 43\(^{4.10+3}\) =\(\overline{........7}\) (tận cùng là 7)

Tương tự ta có 17\(^{17}\) cũng có tận cùng là 7

⇒43\(^{43}\)- 17\(^{17}\) tận cùng là 0, chia hết cho 10

Ta có: \(43^4=***1\)

=>(43⁴)¹⁰= *****1

=>(43⁴)¹⁰.4³=******7

17⁴=***1

=>(17⁴)⁴=*****1

=>(17⁴)⁴.17=*****7

=>A=********7-*****7=********0

=>A chia hết cho 10

a: \(\frac{4^{30}\cdot3^{43}}{2^{53}\cdot27^{15}}\)

\(=\frac{\left(2^2\right)^{30}\cdot3^{43}}{2^{53}\cdot\left(3^3\right)^{15}}\)

\(=\frac{2^{60}}{2^{53}}\cdot\frac{3^{43}}{3^{45}}=\frac{2^7}{3^2}=\frac{128}{9}\)

b: \(\left(-27\right)^{10}\cdot16^{25}\)

\(=-3^{30}\cdot\left(2^4\right)^{25}=-3^{30}\cdot2^{100}\)

\(6^{30}\cdot\left(-32\right)^{15}\)

\(=2^{30}\cdot3^{30}\cdot\left(-1\right)\cdot\left(2^5\right)^{15}=-2^{105}\cdot3^{30}\)

\(\frac{\left(-27\right)^{10}\cdot16^{25}}{6^{30}\cdot\left(-32\right)^{15}}\)

\(=\frac{-2^{100}\cdot3^{30}}{-2^{105}\cdot3^{30}}=\frac{1}{2^5}=\frac{1}{32}\)

c: \(\left(-\frac{2}{17}\right)^3\cdot\left(-\frac{17}{35}\right)^3\cdot\left(\frac{35}{3}\right)^3\)

\(=\left(-\frac{2}{17}\cdot\frac{-17}{35}\cdot\frac{35}{3}\right)^3\)

\(=\left(\frac23\right)^3=\frac{8}{27}\)

e: \(2^2+4^2+\cdots+24^2\)

\(=2^2\left(1^2+2^2+\cdots+12^2\right)\)

\(=4\cdot650=2600\)

a ) \(-5=\frac{-5}{1}< 0\)và \(\frac{1}{63}>0\)

\(\Rightarrow-5< \frac{1}{63}\)