a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)

b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)

- Phần a): phân tích thành các thừa số chung, rồi rút chúng ra ngoài dấu ngoặc, sau đó rút gọn.

- Phần b): phân tích một số thành tích các số, sau đó rút gọn các số giống nhau ở tử và mẫu.

mấy cái này ta chỉ cầm bấm máy tính thôi
\(\dfrac{7.25-49}{7.24+21}=\dfrac{2}{3}\)

\(\dfrac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\dfrac{-3}{2}\)

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

Bạn tự trực tiêu các số bằng nhau nha

Chúc học tốt!

\(\frac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}\)

\(=\frac{1.\left(-1\right).3.2}{-1.2.\left(-1\right).2}\)

\(=\frac{-6}{4}=\frac{-3}{2}\)

Vậy phân số \(\frac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}\) rút gọn là \(\frac{-3}{2}\)

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 !

ừ bài nâng cao mà bạn ơi :)))

\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)

\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)

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,

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

Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?

ĐK: x\(\ne\){-3;0;3}.

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