Bài 1: Thu gọn

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)

d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)

e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)

f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)

g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)

h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)

i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)

k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)

n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)

m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)

p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)

a)Tính: \({\left( {\frac{{ - 1}}{2}} \right)^5};{\left( {\frac{{ - 2}}{3}} \right)^4};{\left( { - 2\frac{1}{4}} \right)^3};{\left( { - 0,3} \right)^5};{\left( { - 25,7} \right)^0}\).

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

Hãy rút ra nhận xét về dấu của luỹ thừa với số mũ chẵn và luỹ thừa với số mũ lẻ của một số hữu tỉ âm.

1 TÍNH

\(a,\left(\frac{-1}{4}\right)^0\)

\(b,\left(-2\frac{1}{3}\right)^2\)

\(c,\left(\frac{4}{5}\right)^{-2}\)

\(d,\left(0,5\right)^{-3}\)

\(e,\left(-1\frac{1}{3}\right)^4\)

\(f,27^3:3^2\)

\(g,\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5\)

\(h,5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3\)

\(i,\left(\frac{1}{3}\right)^{-3}+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right].8\)

Tính:

A=\(\left(1-\frac{2}{3}+\frac{4}{3}\right)-\left(\frac{4}{5}-1\right)+\left(\frac{7}{5}+2\right)\)

B=\(\left(-3+\frac{3}{4}-\frac{1}{3}\right):\left(5+\frac{2}{5}-\frac{2}{3}\right)\)

C=\(\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right)\)\(.\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{11}{12}\right)\)

D=\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{-5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{4}{3}}\)

a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)

b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)

c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)

d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)

g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)