Bt1:thực hiện phép tính
\(\frac{1}{x-1}-\frac{x^3-x}{x^2+1}-\orbr{\begin{cases}\\\end{cases}\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\orbr{\begin{cases}\\\end{cases}}}\)
\(\orbr{\begin{cases}\\\end{cases}}6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\orbr{\begin{cases}\\\end{cases}\frac{3}{2}}\)
Tính hợp lí :\(\orbr{\begin{cases}\\\end{cases}9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right):\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)}\)
\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
\(TÍNH\: GIA\: TRI\: BIEU\: THƯC\: A=\: \: \: \frac{1}{\frac{1}{\frac{100}{1}+}}+\frac{1}{\frac{2}{\frac{49}{2,}+}}+\frac{1}{\frac{3}{\frac{48}{3}+...+}}+...+\frac{1}{\frac{49}{\frac{2}{48}+\:\:}}+\frac{1}{\frac{50}{\frac{1}{50\:}}}\)\(\)
Cho biểu thức:\(A=\orbr{\begin{cases}\left(x-1\right)^2\\3x+\left(x-1\right)^2\end{cases}}-\frac{1-2x^2+4x}{x^3-1}\)\(+\frac{1}{x-1}]\)\(:\frac{x^2+x}{x^3+x}\)
a)Rút gọn biểu thức A
b)Tìm giá trị của x để A>-1
\(\orbr{\begin{cases}\\4\frac{1}{2}:2,5-40\%+\left(-\frac{1}{5}\right)\end{cases}}.\frac{5}{9}\)
\(\left[4\frac{1}{2}:2,5-40\%+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(=\left[\frac{9}{2}\cdot\frac{5}{2}-\frac{2}{5}+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(\left[\frac{45}{4}-\frac{2}{5}+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(=\frac{213}{20}\cdot\frac{5}{9}=\frac{71}{12}\)
xin ti c k
\(\left(0,125+40\%-\frac{3}{40}\right)\):\(\orbr{\begin{cases}\\\end{cases}11\frac{3}{7}+8\frac{1}{2}-\left(\frac{13}{12}-5\frac{4}{7}\right)}\)
Tính A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}\)
A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\frac{50}{50}}\)
A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+\frac{1}{48}+\frac{50}{47}+...+\frac{1}{2}+\frac{1}{50}\right).50}=\frac{1}{50}\)
\(A=\frac{T}{M}\)
\(M=\frac{1}{49}+1+\frac{2}{48}+1+\frac{3}{47}+1+.........+\frac{48}{2}+1+1\)
\(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+.........+\frac{50}{2}+1\)
\(=50.\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+......+\frac{1}{2}+\frac{1}{50}\right)=50.T\)
\(A=\frac{T}{50T}=\frac{1}{50}\)
\(\orbr{\begin{cases}\\\end{cases}\frac{x^2-y^2}{xy}-\frac{1}{x+y}\hept{ }\frac{x^2}{y}-\frac{y^2}{x}\hept{ }\orbr{\begin{cases}\\\end{cases}}}\)
