a: \(A=\dfrac{4^{15}\cdot7^{15}\cdot3^{17}}{4^{32}\cdot3^{16}}=\dfrac{1}{4^{17}}\cdot3\cdot7^{15}=\dfrac{3\cdot7^{15}}{4^{17}}\)

b: \(B=\dfrac{3^{10}+6^2}{5\cdot3^8+20}=\dfrac{3^{10}+3^2\cdot4}{5\left(3^8+4\right)}=\dfrac{3^2\left(3^8+4\right)}{5\left(3^8+4\right)}=\dfrac{9}{5}\)

1)

a) \(\left(-48\right)^3:16^3\)

\(=\left(-48:16\right)^3\)

\(=\left(-3\right)^3\)

\(=-27.\)

b) \(\left(\frac{9}{10}\right)^6:\left(\frac{17}{-20}\right)^6\)

\(=\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)

\(=\left(-\frac{18}{17}\right)^6\)

Chúc em học tốt!

\(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)1.

a, (-48)³:16³

= \(\left(\frac{-48}{16}\right)^3\)

= (-3)³

b,\(\left(\frac{9}{10}\right)^6\):\(\left(\frac{17}{-20}\right)^6\)

= \(\left(\frac{9}{10}:\frac{17}{-20}\right)^6\)

=\(\left(\frac{-18}{17}\right)^6\)

c, \(\left(\frac{-13}{8}\right)^3:\left(\frac{-32}{13}\right)^4\)

= \(\frac{-13^3}{\left(2^3\right)^3}:\frac{\left(-2^5\right)^4}{13^4}\)

= \(\frac{-13^3}{2^9}.\frac{-13^4}{2^{20}}\)

=\(\frac{13^7}{2^{29}}\)

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) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)

\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)

\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)

c) Tìm các số nguyên x,y thỏa mãn

*\(2xy+6x-y=10\)

\(\Leftrightarrow\left(2xy+6x\right)-y-3=10-3=7\)

\(\Leftrightarrow2x\left(y+3\right)-\left(y+3\right)=7\)

\(\Leftrightarrow\left(y+3\right)\left(2x-1\right)=7\)

Lập bảng xét ước nữa là xong.

* \(xy+4x-3y=1\Leftrightarrow\left(xy+4x\right)-3y-12=1-12=-11\)

\(\Leftrightarrow x\left(y+4\right)-\left(3y+12\right)=-11\)

\(\Leftrightarrow x\left(y+4\right)-3\left(y+4\right)=-11\)

\(\Leftrightarrow\left(x-3\right)\left(y+4\right)=-11\)

Lập bảng xét ước nữa là xong.

Mới nhìn vào thấy bài toán hay hay lạ kì.

Thêm một vào bớt một ra

Tức thì bài toán trở nên dễ dàng:

 \(\frac{x}{50}-\frac{x-1}{51}=\frac{x+2}{48}-\frac{x-3}{53}\) 

\(\Leftrightarrow\frac{x}{50}+1-\frac{x-1}{51}-1=\frac{x+2}{48}+1-\frac{x-3}{53}-1\)

\(\Leftrightarrow\left(\frac{x}{50}+1\right)-\left(\frac{x-1}{51}+1\right)=\left(\frac{x+2}{48}+1\right)-\left(\frac{x-3}{53}+1\right)\)

\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}=\frac{x+50}{48}-\frac{x+50}{53}\)

\(\Leftrightarrow\frac{x+50}{50}-\frac{x+50}{51}-\frac{x+50}{48}+\frac{x+50}{53}=0\)

\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)=0\)

Dễ thấy \(\left(\frac{1}{50}-\frac{1}{51}-\frac{1}{48}+\frac{1}{53}\right)\ne0\)

Do đó x + 50 = 0 hay x = -50

bạn ơi ! đề bài

a/ \(5\dfrac{5}{47}+\dfrac{37}{53}+2,7-\dfrac{5}{47}+\dfrac{6}{53}\)

= \(\dfrac{240}{47}-\dfrac{5}{47}+\dfrac{37}{53}+\dfrac{6}{53}+2,7\)

=\(\left(\dfrac{240}{47}-\dfrac{5}{47}\right)+\left(\dfrac{37}{53}+\dfrac{6}{53}\right)+2,7\)

= 5 + \(\dfrac{43}{53}\) + 2,7 = \(\dfrac{4511}{530}\)

b/ \(42\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)-52\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)\)

= \(\left(42\dfrac{1}{6}-52\dfrac{1}{6}\right):\left(-1\dfrac{3}{5}\right)\)

= \(-10:\left(-1\dfrac{3}{5}\right)\) =25/4

a)(2³)¹⁷-2⁴⁸

=2⁵¹-2⁴⁸

=2⁴⁸.(2³-1)

=(2⁴⁸.7)chia hết cho 7

c)=(3³)¹³-(3²)¹⁸

=3³⁹-3³⁶

=3³⁶.(3³-1)

=3³⁶.26

=3³⁵.3.13.2

=(3³⁵.39.2)chia hết cho 39

cho tớ nha thanks