\(\sqrt{\left(-7\right).\left(-14\right)a^4b^2}\)

\(=\sqrt{98a^4b^2}\)

\(=\sqrt{2.49a^4b^2}\)

\(=\sqrt{\left(a^2b.7\sqrt{2}\right)^2}\)

\(=\left|a^2b.7\sqrt{2}\right|\)

\(=a^2\left|b\right|7\sqrt{2}\)

1. \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=\dfrac{-\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}=-\sqrt{a}\)

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

77. listening
78. has finished
79. be recycled
80. goes
81. go
82. talking
83. reading
84. speaking
85. to pass
86. having
87. listening
88. playing
89. to help
90. opening
91. not to make
92. to work
93. writing
94. to buy
95. go
96. to take

54. wouldn't fail
55. doing
56. are installed
57. organizing
58. buys
59. do
60. hadn't known
61. to clean
62. traveling
63. were paid
64. fixing
65. containing
66. would go
67. rains
68. haven't met

Bài tập 5:

a. \(A=\left(2x^2+4x+2\right)+13\)

\(=2\left(x^2+2x+1\right)+13\)

\(=2\left(x+1\right)^2+13\ge13\)

Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy \(MinA=13\Leftrightarrow x=-1\)

b. \(B=3x^4+x^2+10\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(MinB=10\Leftrightarrow x=0\)

c. \(B=4x^2+3x+5\)

\(=\left(4x^2+2.\dfrac{3}{4}.2x+\dfrac{9}{16}\right)-\dfrac{9}{16}+5\)

\(=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}\ge\dfrac{71}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow2x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{8}\)

Vậy \(MinC=\dfrac{71}{16}\Leftrightarrow x=-\dfrac{3}{8}\)

d. \(D=\left(x+5\right)\left(x-1\right)\left(x^2+4x+5\right)\)

\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)

\(=\left[\left(x^2+4x\right)-5\right]\left[\left(x^2+4x\right)+5\right]\)

\(=\left(x^2+4x\right)^2-25\ge-25\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+4x=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(MinD=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Bài tập 6:

a. \(A=\left(x-1\right)^3-\left(x+1\right)^3-3\left(2x^2-1\right)\)

\(=\left(x^3-3x^2+3x-1\right)-\left(x^3+3x^2+3x+1\right)-\left(6x^2-3\right)\)

\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1-6x^2+3\)

\(=-12x^2+1\)

b. \(B=4\left(x-2\right)\left(x+2\right)-\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-2\right)-2\left(x^2+1\right)\)

\(=4\left(x^2-4\right)-\left(x^3-1\right)+\left(x^3-2x^2\right)-\left(2x^2+2\right)\)

\(=4x^2-16-x^3+1+x^3-2x^2-2x^2-2\)

\(=-17\)

c. \(C=\left(y-x+1\right)^2+2\left(x-y\right)\left(1+y-x\right)+\left(x-y\right)^2\)

\(=\left(y-x+1\right)^2+2\left(x-y\right)\left(y-x+1\right)+\left(x-y\right)^2\)

\(=\left[\left(y-x+1\right)+\left(x-y\right)\right]^2\)

\(=\left(y-x+1+x-y\right)^2\)

\(=1\)

\(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{2}{20}-\dfrac{2}{30}-\dfrac{2}{42}-...-\dfrac{2}{240}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{2.\left(4-1\right)}{16}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{2.3}{16}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\)

\(\dfrac{x}{2008}=1\)

\(x=2008\)

\(A=\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)

\(=\dfrac{2}{4.9}+\dfrac{2}{9.14}+\dfrac{2}{14.19}+...+\dfrac{2}{504.509}\)

\(=\dfrac{2.5}{5.4.9}+\dfrac{2.5}{5.9.14}+\dfrac{2.5}{5.14.19}+...+\dfrac{2.5}{5.504.509}\)

\(=\dfrac{2}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{504.509}\right)\)

\(=\dfrac{2}{5}\left(\dfrac{9-4}{4.9}+\dfrac{14-9}{9.14}+\dfrac{19-14}{14.19}+...+\dfrac{509-504}{504.509}\right)\)

\(=\dfrac{2}{5}\left(\dfrac{1}{4}-\dfrac{1}{509}\right)\)

\(=\dfrac{2}{5}.\dfrac{509-4}{4.509}\)

\(=\dfrac{2}{5}.\dfrac{505}{4.509}\)

\(=\dfrac{101}{2.509}\)

\(=\dfrac{101}{1018}\)

\(B=\dfrac{1}{10.9}+\dfrac{1}{18.13}+\dfrac{1}{26.17}+...+\dfrac{1}{802.405}\)

\(=\dfrac{1}{2.5.9}+\dfrac{1}{2.9.13}+\dfrac{1}{2.13.17}+...+\dfrac{1}{2.401.405}\)

\(\Rightarrow4B=4\left(\dfrac{1}{2.5.9}+\dfrac{1}{2.9.13}+\dfrac{1}{2.13.17}+...+\dfrac{1}{2.401.405}\right)\)

\(=\dfrac{4}{2.5.9}+\dfrac{4}{2.9.13}+\dfrac{4}{2.13.17}+...+\dfrac{4}{2.401.405}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{401.405}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{9-5}{5.9}+\dfrac{13-9}{9.13}+\dfrac{17-13}{13.17}+...+\dfrac{405-401}{401.405}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)

\(=\dfrac{1}{2}.\dfrac{81-1}{405}\)

\(=\dfrac{1}{2}.\dfrac{80}{405}\)

\(=\dfrac{40}{405}\)

\(\Rightarrow B=\dfrac{1}{4}.\dfrac{40}{405}=\dfrac{2}{81}\)

2. Jane said that Alex and Jane had given their old clothes to the poor.

3. David said that the show had ended and the new one was starting in a few minutes.

4. Mike told his mother that a lady had been there to sell cupcakes the previous night.

5. Jim said that their class would start the following week.

a) \(x^3+y^3+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

b) Ta có:

\(VT=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)

\(=\left[\left(a-b\right)^3+\left(b-c\right)^3\right]+\left(c-a\right)^3\)

\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)

\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3\)

\(=\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-\left(a-c\right)^3\)

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)=VP\) (đpcm)