\(\left(x+\dfrac{2}{3}\right)\left(\dfrac{1}{4}-x\right)>0\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}>0\\\dfrac{1}{4}-x>0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>-\dfrac{2}{3}\\x< \dfrac{1}{4}\end{matrix}\right.\)

Để có từ khiêm tốn, em sẽ gõ như thế nào theo kiểu Telex

A. khieem toons                 B. khiem toonr                     C. khieem toonr                     D. khieem toonf

Để có từ quê hương, em sẽ gõ như thế nào theo kiểu Telex?

A. quee huong                      B. quee huwowng                       C. que huwowng

\(1,\dfrac{4\sqrt{x}-x-4}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{-\left(x-4\sqrt{x}+4\right)}{\sqrt{x^2}-2^2}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}=\dfrac{2-\sqrt{x}}{\sqrt{x}+2}\)

\(2,\dfrac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\left(dk:x,y\ge0\right)\\ =\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

\(3,\dfrac{x-9}{x\sqrt{x}-27}\left(dk:x\ge0\right)\\ =\dfrac{\sqrt{x^2}-3^2}{\sqrt{x^3}-3^3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)

\(\dfrac{x+a\sqrt{x}}{a\sqrt{x}}\left(dk:x>0,a\ne0\right)\\ =\dfrac{\sqrt{x^2}+a\sqrt{x}}{a\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+a\right)}{a\sqrt{x}}=\dfrac{\sqrt{x}+a}{a}\)

\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)

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

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)

Bạn An gặp tình huống Bạo lực mạng.

\(12.120:\left[3\left(x-25\right)\right]=10\\ \Rightarrow1440:\left[3\left(x-25\right)\right]=10\\ \Rightarrow3\left(x-25\right)=144\\ \Rightarrow x-25=48\\ \Rightarrow x=73\)

\(4545-3\left(x-2\right)=12\\ \Rightarrow3\left(x-2\right)=4533\\ \Rightarrow x-2=1511\\ \Rightarrow x=1513\)