Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-4.2.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{358}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}+2.\dfrac{1}{297}.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-4.2.\dfrac{1}{297}\)
\(A=0+0+\dfrac{-8}{297}\)
\(A=\dfrac{-8}{297}\)
Chúc bạn học tốt!!!
A= \(\dfrac{1}{358}\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
A= \(\dfrac{7}{358}+\dfrac{1}{358.297}-\dfrac{8}{297}+\dfrac{2}{358.297}-\dfrac{7}{358}-\dfrac{3}{358.297}\)
A= \(-\dfrac{8}{297}\)
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)
Câu hỏi của Kurosaki Akatsu - Toán lớp 8 - Học toán với OnlineMath
Câu hỏi của Kurosaki Akatsu - Toán lớp 8 - Học toán với OnlineMath