\(\dfrac{1}{20}+\dfrac{3}{5}=\dfrac{1}{20}+\dfrac{12}{20}=\dfrac{13}{20}\\ \dfrac{4}{5}\times\dfrac{20}{9}=\dfrac{4\times4\times5}{5\times9}=\dfrac{4\times4}{9}=\dfrac{16}{9}\\ \dfrac{4}{5}-\dfrac{5}{8}=\dfrac{32}{40}-\dfrac{2}{40}=\dfrac{7}{40}\\ \dfrac{1}{2}:\dfrac{5}{8}=\dfrac{1}{2}\times\dfrac{8}{5}=\dfrac{4\times2}{2\times5}=\dfrac{4}{5}\)

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne\pm1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)

\(b,A=3\\ \Leftrightarrow\dfrac{x+2}{2x+1}=3\\ \Leftrightarrow6x+3=x+2\\ \Leftrightarrow5x+1=0\\ \Leftrightarrow x=-\dfrac{1}{5}\left(tm\right)\)

\(c,\dfrac{1}{A}=\dfrac{2x+1}{x+2}=\dfrac{2x+4-3}{x+2}=\dfrac{2\left(x+2\right)-3}{x+2}=2-\dfrac{3}{x+2}\)

Để `1/A` là số nguyên thì `3/(x+2)` nguyên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Ta có bảng:

Vậy \(x\in\left\{-5;-3\right\}\)

\(a,2+\dfrac{1}{4}\times\dfrac{4}{3}=2+\dfrac{1}{3}=\dfrac{6}{3}+\dfrac{1}{3}=\dfrac{6+1}{3}=\dfrac{7}{3}\)

có thưởng ko nhỉ :>

\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)

bạn đăng thành câu hỏi mới, nó chx sửa đc đâu

\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{x-1}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{x-1}{2x+1}\\ =\dfrac{x+2}{\left(x+1\right)\left(2x+1\right)}\)

Đề sai r bn

Gọi pt đt đó là y=ax+b

 Để đt đó đi qua P(0;-1) thì -1=0a+b⇒-1=b(1)

 Để đt đó đi qua Q(3;8) thì 8=3x+b(2)

Từ (1), (2) ta có hpt:
\(\left\{{}\begin{matrix}b=-1\\8=3a+b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-1\\8=3a+\left(-1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-1\\a=3\end{matrix}\right.\)

Vậy pt đường thẳng đó là: y=3x-1

a, Thay x=1, y=`-1/2`, z=-2 vào đơn thức ta có:
\(-\dfrac{1}{3}x^2y^3z^3=-\dfrac{1}{3}.1^2.\left(-\dfrac{1}{2}\right)^3.\left(-2\right)^3=-\dfrac{1}{3}.1.\left(-\dfrac{1}{8}\right).\left(-8\right)=-\dfrac{1}{3}\)

b, \(5xy^2.0,7y^4z.4xyz=14x^2y^7z^2\)

Hệ số: 14

Biến: x²y⁷z²

Bậc: 11