\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}\)

= \(\frac{\left(-125\right).40.64}{135.16384.1}\)

= \(\frac{\left(-5000\right).64}{2211840}\)

= \(\frac{-320000}{2211840}\)

= \(\frac{-125}{864}\)

Lưu ý: Kết quả và cách làm chưa chắc đúng. Mong bạn cẩn thận. Nếu đúng, chọn đúng cho mình nhé.

\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\frac{\left(-5\right)^3.40.4^3}{5.27.4^7}=\frac{\left(-5\right)^3.2}{3^3.4^3}=\frac{\left(-5\right)^3}{3^3.2^5}\)

có gì sai bỏ qua

\(\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.1}=\dfrac{\left(-15\right).40.64}{135.\left(-2\right)^{14}}=\dfrac{\left(-5\right).3.5.8.64}{15.9.\left(-2\right)^{14}}=\dfrac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}=\dfrac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\dfrac{5}{9.\left(-2\right)^5}\)

\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}=\frac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}=\frac{5.8.3.\left(-5\right).64}{15.9.\left(-2\right)^{14}}=\frac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}=\frac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\frac{5}{9.\left(-2\right)^5}\)

a) = -320000/2211840 = -125/864

= ((-5)^3 . 5 . 2^3 . 2^6] / ( 5. 3^3 . 2^14 .1)

= ( (-5

\(\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(100\right)^0}=\dfrac{\left(-5\right)^3.2^3.5.2^6}{3^2.5.2^{14}}\)

* giải thích : \(100^0\)= 1 nên mik ko viết nữa

\(\left(-2\right)^{14}=\left(2\right)^{14}\)

ta có : \(\dfrac{\left(-5\right)^3.2^9.5}{3^2.2^{14}.5}=\dfrac{-125}{9.32}=\dfrac{-125}{288}\)

tik mik nhé

\(\frac{-5^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\left(-100\right)^0}=\frac{-125\cdot2^3\cdot5\cdot\left(2^2\right)^3}{5\cdot27\cdot2^{14}\cdot1}=\frac{-125\cdot2^6}{27\cdot2^{11}}=\frac{-125}{27\cdot2^5}=\frac{-125}{864}\)

\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}\)\(=\frac{\left(-5\right)^3.5.2^3.2^6}{3^3.5.2^{14}.1}\)\(=\frac{-125}{864}\)

\(a,\dfrac{2929-101}{2.1919+404}=\dfrac{29.101-101.1}{2.19.101+4.101}\)

\(=\dfrac{101\left(29-1\right)}{101\left(2.19+4\right)}\)

\(=\dfrac{101.29}{101.42}\)

\(=\dfrac{28}{42}=\dfrac{2}{3}\)

\(b,\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}\)

\(=\dfrac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}\)

\(=\dfrac{5.8.3.\left(-5\right).64}{5.3.9.\left(-2\right)^{14}}\)

\(=\dfrac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}\)

\(=\dfrac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\dfrac{5}{9.\left(-2\right)^5}\)