Thánh làm

\(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{2^{15}\cdot5^8-2^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^{10}\cdot5^{18}-10^9\right)\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^2\cdot10^8-10^9\right)\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^2-10\right)\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(4-10\right)\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-6\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot2\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot2^6\cdot10^8}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot64\cdot10^8}{2^{16}\cdot5^7+20^8}\)

Bài 2:

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

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

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

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

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

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

\(=\dfrac{6x^3+9x^2+2x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{\left(2x^2+3x\right)\left(1+3x\right)}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+6x^3+3x+9x^2}\)

\(=\dfrac{6x^3+11x^2+6x+3}{11x^2+6x^3+3x}\)

Bài 1:

\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)

\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7-x}{x-7}\)

\(=\dfrac{-\left(x-7\right)}{x-7}\)

\(=-1\)

Nhờ thầy phynit đổi tên thôi, anh chịu :3

Câu hỏi của Heo Sun - Hóa học lớp 9 | Học trực tuyến