We haven't called each other for a long time 

a. ... = \(\sqrt{3}\left(\sqrt{3}-1\right)-\sqrt{15}\left(\sqrt{3}-1\right)\) = \(\sqrt{3}\left(1-\sqrt{5}\right)\left(\sqrt{3}-1\right)\)

b. ... = \(\sqrt{1-a}+\sqrt{1-a^2}\left(-1< a< 1\right)\)  \(=\left(\sqrt{1+a}+1\right)\sqrt{1-a}\)

c. ... = \(a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\) = \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2\)

d. ... = \(x-y+y\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)

\(X=\dfrac{125-65}{10}=6\)

\(\dfrac{12}{5}=\dfrac{12.12}{5.12}=\dfrac{144}{60};\dfrac{19}{12}=\dfrac{19.5}{12.5}=\dfrac{95}{60}\)  \(\Rightarrow\dfrac{12}{5}>\dfrac{19}{12}\)

sin A = \(\dfrac{CF}{AC};sinB=\dfrac{CF}{BC}\)  . Suy ra : \(\left\{{}\begin{matrix}\dfrac{BC}{sinA}=\dfrac{BC.AC}{CF}\\\dfrac{AC}{sinB}=\dfrac{BC.AC}{CF}\end{matrix}\right.\) \(\Rightarrow\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)

CMTT : \(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\) . Suy ra đpcm 

a. ... = \(2x^2+x-x^3-2x^2+x^3-x+3=3\)

b. ... = \(4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)

\(\dfrac{102}{144}=\dfrac{102:6}{144:6}=\dfrac{17}{24}\)

\(3\left(-\dfrac{4}{3}\right)^2-\dfrac{7}{3}-0,8.1=3.\dfrac{16}{9}-\dfrac{7}{3}-\dfrac{4}{5}\)  = \(\dfrac{16}{3}-\dfrac{7}{3}-\dfrac{4}{5}=3-\dfrac{4}{5}=\dfrac{11}{5}\)

\(\dfrac{9}{25}-\dfrac{4}{5}.\dfrac{64}{25}=\dfrac{9}{25}-\dfrac{256}{125}=\dfrac{-211}{125}\)

B/t A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\4-x>0\end{matrix}\right.\)  \(\Leftrightarrow3\le x< 4\)

B/t B có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\x^2-4x+4>0\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x>1\\\left(x-2\right)^2>0\end{matrix}\right.\)  \(\Leftrightarrow2\ne x>1\)

1. This is my friend , Lan.

2. Your library is new.

3. Is your library new ?

4. His playground is large.

5. Her ball is under the bed.