a, ta có \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)=\(\frac{2.\left(-13\right).\left(-3\right).\left(-3\right).2.\left(-5\right)}{\left(-3\right).2.2.\left(-5\right).\left(-2\right).\left(-13\right)}\)

rút gọn đi còn: \(\frac{-3}{-2}\)=\(\frac{3}{2}\)

15×8+15×4/12×3

= 15×(8+4)/12×3

=15×12/12×3

=15/3

=5

a, Ta có cách làm sau

\(\dfrac{15.8+15.4}{12.3}=\dfrac{15.\left(8+4\right)}{12.3}=\dfrac{15.12}{12.3}=\dfrac{3.5.12}{12.3}\)

\(=\dfrac{5}{1}=5\)

b,

\(\dfrac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}\)=\(\dfrac{2.\left(-13\right).\left(-3\right).\left(-3\right).\left(-2\right).\left(-5\right)}{-3.2.2_{ }.\left(-5\right).\left(-2\right).\left(-13\right)}\)=\(\dfrac{-3}{2}\)

\(=\left[\dfrac{a^6b^3}{c^3d^6}\cdot\dfrac{ac^4}{b^2d^3}\right]:\left[\dfrac{a^8b^8}{c^4d^{12}}\cdot\dfrac{c^3}{b^9d^3}\right]\)

\(=\dfrac{a^7b^3c^4}{c^3d^9b^2}:\dfrac{a^8}{bcd^{15}}\)

\(=\dfrac{a^7bc}{d^9}\cdot\dfrac{bcd^{15}}{a^8}=\dfrac{d^6\cdot b^2\cdot c^2}{a}\)

a, \(\frac{7.25-49}{7.24+21}\)

\(=\frac{7.25-7.7}{7.24+7.3}=\frac{7\left(25-7\right)}{7\left(24+3\right)}=\frac{18}{27}=\frac{2}{3}\)

b, \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{2.\left(-13\right).\left(-3\right).\left(-3\right).\left(-5\right).\left(-2\right)}{\left(-3\right).2.2.\left(-5\right).\left(-13\right).\left(-2\right)}\)

\(=\frac{-3}{2}\)

DỄ LẮM !

a) \(\frac{7\cdot25-49}{7\cdot24+21}=\frac{7\cdot25-7\cdot7}{7\cdot24+7\cdot3}=\frac{7\left(25-7\right)}{7\left(24+3\right)}\)

=\(\frac{18}{27}=\frac{2}{3}\)

b) \(\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\frac{-1\cdot2\cdot13\cdot3\cdot3\cdot5\cdot2}{1\cdot3\cdot2\cdot2\cdot5\cdot13\cdot2}=-\frac{3}{2}\)

Chúc bn học tốt !

\(a,\frac{2.25-49}{2.24+21}=\frac{2.25-7.7}{7.24+7.3}=\frac{7.\left(25-7\right)}{7.\left(24+3\right)}=\frac{2}{3}\)

\(b,\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{2.\left(-1\right).13.3.3.5.2}{1.3.2.2.5.13.2}\frac{-3}{2}\)

\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right) \left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4