Question

Bài 1: Tính;

a/ M = \(\dfrac{4^6.9^5+120.6^9}{8^4.3^{12}-6^{11}}\)

b/ N = \(\left(-\dfrac{3}{4}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(2\dfrac{1}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\)

Bài 2:

a, Tìm 2 số khác 0 biết rằng tổng, hiệu, tích của chúng tỉ lệ với 5; 1; 12

b, Cho: x, y, z khác 0 và x² = yz. Chứng minh rằng: \(\dfrac{x^2+y^2}{x^2+z^2}=\dfrac{y}{z}\)

Answer 1

Answer 2

Bài 1: 

a: \(M=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)

b: \(N=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)

\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)

\(=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)