\(a,\frac{30x^3}{11y^2}.\frac{121y^5}{25x}\)

\(=>\frac{30x^3.121y^5}{11y^2.25x}=\frac{6x^2.11y^3}{5}=\frac{66x^2.y^3}{5}\)

\(b,\frac{24y^5}{7x^2}.\frac{-21x}{12y^3}\)

\(=>\frac{24y^5.\left(-21\right)x}{7x^2.12y^3}=\frac{2y^2.\left(-3\right)}{x}=-\frac{6y^2}{x}\)

\(c,\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)\)

\(=>\frac{-18y^3.\left(-15\right)x^2}{25x^4.9y^3}=\frac{-2.\left(-3\right)}{5x^2}=\frac{6}{5x^2}\)

\(d,\frac{3x^2}{2y}.\frac{1}{4y}.\frac{5}{3y}\)

\(=>\frac{3x^2.1.5}{2y.4y.3y}=\frac{15x^2}{24y^3}=\frac{5x^2}{8y^3}\)

\(e,\frac{2x}{3}.\frac{x+1}{2x}\)

\(=>\frac{2x\left(x+1\right)}{3.2x}=\frac{x+1}{3}\)

\(g,\frac{5-x}{x-3}.\frac{2}{3}.\frac{x}{4}\)

\(=>\frac{2x\left(5-x\right)}{3.4\left(x-3\right)}=\frac{10x-2x^2}{12\left(x-3\right)}=\frac{10x-2x^2}{12x-9}\)

\(\dfrac{10x^3}{11y^2}.\dfrac{121y^5}{25x}=\dfrac{1210x^3y^5}{275xy^2}=\dfrac{22x^2y^3}{5}\)

a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)

\(=\dfrac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\)

b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)

c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)

d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)

A. Cách B sai vì 5 : 2/5 thì ko thể nào = 25 đc.

\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

A= 4/7.

Biết có cái

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)

c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)

d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)

\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)

\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)

e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)

\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)

\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)

f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)

\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)

\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)

g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)

\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)

\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)

\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)

\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)

\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)

\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)

\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)

\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)

a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729

b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)

c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)

\(=\dfrac{2^3}{4^4}\cdot7\)

\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)

d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)

\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)

\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)

e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)

\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)

\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)

f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)

\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)

g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)

\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)

\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)