Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$

$=(x-1)(x^2+3x+1)$

III:

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

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

4) \(a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2\)

5) \(2x-\sqrt{x}-3=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)\)

6) \(6a^2-5a\sqrt{b}-b=\left(a-\sqrt{b}\right)\left(6a+\sqrt{b}\right)\)

7) \(x-2\sqrt{x-1}-y^2=\left(\sqrt{x-1}-1\right)^2-y^2=\left(\sqrt{x-1}-1-y\right)\left(\sqrt{x-1}-1+y\right)\)

II:

2.8) ĐKXĐ: \(x\ge2\)

2.9: ĐKXĐ: \(\left[{}\begin{matrix}x< \dfrac{1}{2}\\\dfrac{1}{2}< x\le1\end{matrix}\right.\)

2.10: ĐKXĐ: \(x\ne0\)

2.11: ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge3\end{matrix}\right.\)

1 It was silly to lend him money

2 Meeting the president is exciting

3 While we were on the summer holiday, we were so happy

4 While they were playing, they cried twice

5 Writing good composition in English was not easy

26 It was silly to lend him money
27 Meeting the president is exciting
28 While we were in the summer holiday, we were happy
29 While the play was being played, they cried twice
30 Writing good compositions in English was not easy

26 It was silly to lend him money

27 Meeting the president is exciting

28 While we were having our summer holiday, we were so happy

29 While they were seeing the play, they cried twice

30 Writing good compositions in English was not easy

18.

Áp dụng BĐT quen thuộc: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\) ta có:

\(\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}\ge\dfrac{2}{1+\sqrt{a^3b^3}}\) ; \(\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge\dfrac{2}{1+\sqrt{abc^4}}\)

Cộng vế:

\(\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge2\left(\dfrac{1}{1+\sqrt{a^3b^3}}+\dfrac{1}{1+\sqrt{abc^4}}\right)\ge2\left(\dfrac{2}{1+\sqrt[4]{a^4b^4c^4}}\right)\)

\(\Rightarrow\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}+\dfrac{1}{1+abc}\ge\dfrac{4}{1+abc}\)

\(\Rightarrow\dfrac{1}{1+a^3}+\dfrac{1}{1+b^3}+\dfrac{1}{1+c^3}\ge\dfrac{3}{1+abc}\) (đpcm)

19.

Biến đổi tương đương:

\(\Leftrightarrow\left(a^2+b^2\right)xy+ab\left(x^2+y^2\right)\ge\left(a^2+b^2+2ab\right)xy\)

\(\Leftrightarrow\left(a^2+b^2\right)xy+ab\left(x^2+y^2\right)\ge\left(a^2+b^2\right)xy+2abxy\)

\(\Leftrightarrow ab\left(x^2+y^2\right)-2abxy\ge0\)

\(\Leftrightarrow ab\left(x^2+y^2-2xy\right)\ge0\)

\(\Leftrightarrow ab\left(x-y\right)^2\ge0\)

Ta có :

\(AH^2=AB^2+BH^2\left(1\right)\) (Δ ABH vuông tại H)

\(AH^2=AC^2+CH^2\left(2\right)\) (Δ ACH vuông tại H)

\(\left(1\right),\left(2\right)\Rightarrow AB^2+BH^2=AC^2+CH^2\)

\(\Rightarrow CH^2=AB^2+BH^2-AC^2\)

\(\Rightarrow CH^2=81+676-121=636\)

\(\Rightarrow CH=\sqrt[]{636}=\sqrt[]{4.159}=2\sqrt[]{159}\left(cm\right)\)

Vì AH là đường cao của tam giác ABC nên AH \(\perp\) BC \(\equiv\) H

⇒ \(\Delta\) AHB  \(\perp\)  \(\equiv\) H \(\Rightarrow\) AB > BH ⇒  9 cm > 26 cm vô lý

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                                         2 là em xem lại đề bài của em