Với x >= 0 ; x khác 1 

\(A=\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{\sqrt{x}-1}-x-1\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}-2-x-1=\frac{\sqrt{x}+2}{\sqrt{x}+1}-x-3\)

\(=\frac{\sqrt{x}+2-\left(x+3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\frac{\sqrt{x}+2-x\sqrt{x}-x-3\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\frac{-x-x\sqrt{x}-2\sqrt{x}-1}{\sqrt{x}+1}\)

Quang Trung - Nguyễn Huệ là 1 bạn ạ 

-.-

quang trung và nguyễn huệ là 1 người mà

Gọi giá tiền máy giặt, lò vi sóng lần lượt là a ; b ( a ; b > 0 ) 

Theo bài ra ta có hệ \(\hept{\begin{cases}a+b=21\\\frac{15a}{100}+\frac{10b}{100}=21-18,3=2,7\end{cases}}\Leftrightarrow\hept{\begin{cases}a=12\\b=9\end{cases}}\)(tm) 

Vậy ... 

đk : x >= 2 

\(3\sqrt{x-2}+2x=\sqrt{x+6}+6\)

\(\Leftrightarrow3\sqrt{x-2}-3+2x-6-\left(\sqrt{x+6}-3\right)=0\)

\(\Leftrightarrow\frac{9\left(x-2\right)-9}{3\sqrt{x-2}+3}+2\left(x-3\right)-\frac{x+6-9}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow\frac{9x-27}{3\sqrt{x-2}+3}+2\left(x-3\right)-\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{9}{3\sqrt{x-2}+3}+2-\frac{1}{\sqrt{x+6}+3}\ne0\right)=0\Leftrightarrow x=3\)(tmđk) 

Ta có : \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

\(T=\frac{a^{2021}+b^{2021}+c^{2021}}{\left(a+b+c\right)^{2021}}=\frac{b^{2021}+b^{2021}+b^{2021}}{\left(b+b+b\right)^{2021}}=\frac{3b^{2021}}{\left(3b\right)^{2021}}=\frac{3}{3^{2021}}=\frac{1}{3^{2020}}\)

- Oxit bazo:

+, Li₂O : Liti oxit

+, Na₂O : Natri oxit

+, MgO: Magie oxit

+, Al₂O₃ : Nhôm oxit

- Oxit axit:

+, CO: cacbon monooxit

+, CO₂ : cacbon đioxit

+, NO: nitơ monooxit

+, N₂O₅ : đinitơ pentaoxit

+, N₂O₃ : đinitơ trioxit

Ta có \(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}=2\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}=4\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=4-\dfrac{2}{\sqrt{ab}}\)

Khi đó P = \(\dfrac{1}{\sqrt{ab}}\left(4-\dfrac{2}{\sqrt{ab}}\right)=-2\left(\dfrac{1}{\sqrt{ab}}-1\right)^2+2\le2\)

Dấu "=" khi a = b = 1 

20000 hoặc 40000

20 000 đồng.