\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)(đpcm)

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

\(\Rightarrow\) Đpcm.

\(\dfrac{1}{x}-\dfrac{1}{x+1}\) MTC: \(x\left(x+1\right)\)

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

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

\(=\dfrac{1}{x\left(x+1\right)}\)

\(\Rightarrow dpcm\)

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

a) \(2x^2y^3.\dfrac{1}{4}xy^3\left(-3\right)xy\)

\(=\left(-3.2.\dfrac{1}{4}\right)x^4y^7\)

\(=\dfrac{-3}{2}x^4y^7\)

\(\Rightarrow Hệ\) số: \(\dfrac{-3}{2}\)

Phần biến: \(x^4y^7\)

b) \(\left(-2x^3y\right)^2.xy^2.\dfrac{1}{5}y^5\)

\(=\dfrac{4}{5}x^7y^9\)

\(\Rightarrow Phần\) biến: \(x^7y^9\)

Hệ số: \(\dfrac{4}{5}.\)

a/ \(2x^2y^3\cdot\dfrac{1}{4}xy^3\left(-3xy\right)\)

\(=\left[2\cdot\dfrac{1}{4}\cdot\left(-3\right)\right]\left(x^2.x.x\right)\left(y^3.y^3.y\right)\)

\(=-\dfrac{3}{2}x^4y^7\)

Phần biến: \(x^4y^7\)

Hệ số: \(-\dfrac{3}{2}\)

b/ \(\left(-2x^3y\right)^2\cdot xy^2\cdot\dfrac{1}{5}y^5=4x^6y^2\cdot xy^2\cdot\dfrac{1}{5}y^5\) \(=4\cdot\dfrac{1}{5}\left(x^6\cdot x\right)\left(y^2\cdot y^2\cdot y^5\right)=\dfrac{4}{5}x^7y^9\)

Phần biến: \(\dfrac{4}{5}\)

Hệ số: \(x^7y^9\)

a: \(\dfrac{xy^2}{xy-y}=\dfrac{y\cdot xy}{y\cdot\left(x-1\right)}=\dfrac{xy}{x-1}\)

=>Hai phân thức này bằng nhau

b: \(\dfrac{xy+y}{x}=\dfrac{y\left(x+1\right)}{x}\)

\(\dfrac{xy+x}{y}=\dfrac{x\left(y+1\right)}{y}\)

Vì \(\dfrac{y\left(x+1\right)}{x}\ne\dfrac{x\left(y+1\right)}{y}\)

nên hai phân thức này không bằng nhau

c: \(\dfrac{-6}{4y}=\dfrac{-6:2}{4y:2}=\dfrac{-3}{2y}\)

\(\dfrac{3y}{-2y^2}=\dfrac{-3y}{2y^2}=\dfrac{-3y}{y\cdot2y}=\dfrac{-3}{2y}\)

Do đó: \(\dfrac{-6}{4y}=\dfrac{3y}{-2y^2}\)

=>Hai phân thức này bằng nhau

`a, (xy^2)/(xy+y) = (xy^2)/(y(x+1))`

`=(xy)/(x+1)`

Vậy `2` cặp phân thức bằng nhau.

`b, (xy-y)/x = (y(x-1))/x = (y^2(x-1))/(xy)`

`(xy-x)/y = (x(y-1))/y = (x^2(y-1))/(xy)`

Vậy `2` đa thức không bằng nhau

a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)

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

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

Khôi Bùi , DƯƠNG PHAN KHÁNH DƯƠNG, Mysterious Person, Phạm Hoàng Giang, Phùng Khánh Linh, TRẦN MINH HOÀNG, Dũng Nguyễn, Nhã Doanh, hattori heiji, ...

a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)

\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)

\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)

b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)

\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)

Bạn cần làm gì với biểu thức này?