Theo giả thiết, ta có: 

\(a+b+c=2p\)

\(b+c=2p-a\)

\(\Rightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)

\(b^2+2bc+c^2=4p^2-4pa+a^2\)

Do đó:

\(VT=2bc+b^2+c^2-a^2\)

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

\(=\left(4p^2-4pa+a^2\right)-a^2\)

\(=4p^2-4pa+a^2-a^2\)

\(=4p^2-4pa\)

\(=4p\left(p-a\right)=VP\) (đpcm)

\(\left(x+2\right)\sqrt{x-3}=0\)  \(\left(1\right)\)

Điều kiện: \(x-3\ge0\Leftrightarrow x\ge3\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\sqrt{x-3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

So với điều kiện thấy \(x=3\) thỏa mãn.

Vậy phương trình có tập nghiệm là: \(S=\left\{3\right\}.\)

Bài 2:
1. don't
2. could
3. Shall
4. Let's
5. Why
Bài 3:
1. c
2. a
3. d
4. e
5. b
Bài 4:
2. I'm afraid that I can't come to your party.
3. I'm sorry. I'll be a bit late.
4. I guess I won the competition!

1. A collected /id/ B arranged /d/ C climbed /d/ D carved /d/

2. A collage /a:/ B gardening /a:/ C fragile /æ/ D glass /a:/

3. A eggshell /e/ B pottery /ə/ C many /e/ D exercise /e/

\(2\dfrac{7}{11}+\dfrac{4}{11}\times\dfrac{2}{11}+\dfrac{4}{11}\times\dfrac{9}{11}\)

\(=\dfrac{29}{11}+\dfrac{4}{11}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)

\(=\dfrac{29}{11}+\dfrac{4}{11}\times\dfrac{11}{11}\)

\(=\dfrac{29}{11}+\dfrac{4}{11}\times1\)

\(=\dfrac{29}{11}+\dfrac{4}{11}\)

\(=\dfrac{33}{11}\)

\(=3\)

\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1}\)

\(=\dfrac{1}{50}-\dfrac{50-49}{50.49}-\dfrac{49-48}{49.48}-...-\dfrac{2-1}{2.1}\)

\(=\dfrac{1}{50}-\left(\dfrac{1}{49}-\dfrac{1}{50}\right)-\left(\dfrac{1}{48}-\dfrac{1}{49}\right)-...-\left(1-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{50}-\dfrac{1}{49}+\dfrac{1}{50}-\dfrac{1}{48}+\dfrac{1}{49}-...-1+\dfrac{1}{2}\)

\(=\dfrac{1}{50}+\dfrac{1}{50}-1\)

\(=\dfrac{1+1-50}{50}\)

\(=\dfrac{-24}{25}\)

2. bought

3. saw

4. were

5. did

6. jumped

7. rode

8. had

A)\(\left(\dfrac{3}{15}\right)^{15}:\left(0,36\right)^5\)

\(=\left(\dfrac{1}{5}\right)^{15}:\left(\dfrac{36}{100}\right)^5\)

\(=\dfrac{1}{5^{15}}:\left(\dfrac{36}{100}\right)^5\)

\(=\dfrac{1}{\left(5^3\right)^5}:\left(\dfrac{36}{100}\right)^5\)

\(=\dfrac{1}{125^5}:\left(\dfrac{36}{100}\right)^5\)

\(=\left(\dfrac{1}{125}\right)^5:\left(\dfrac{36}{100}\right)^5\)

\(=\left(\dfrac{1}{125}:\dfrac{36}{100}\right)^5\)

\(=\left(\dfrac{1}{125}.\dfrac{100}{36}\right)^5\)

\(=\dfrac{1}{45^5}\)

B) \(\left(1-\dfrac{1}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\)

\(=\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\)

\(=\left(\dfrac{2}{3}\right)^8:\left[\left(\dfrac{2}{3}\right)^2\right]^3\)

\(=\left(\dfrac{2}{3}\right)^8:\left(\dfrac{2}{3}\right)^6\)

\(=\left(\dfrac{2}{3}\right)^2\)

\(=\dfrac{4}{9}\)

\(2x^2+y^2+z^2+2xy-2xz-yz-y+3z+5=0\)

\(\Leftrightarrow4x^2+2y^2+2z^2+4xy-4xz-2yz-2y+6z+10=0\)

\(\Leftrightarrow\left(4x^2+y^2+z^2+4xy-2yz-4xz\right)+\left(y^2-2y+1\right)+\left(z^2+6z+9\right)=0\)

\(\Leftrightarrow\left(2x+y-z\right)^2+\left(y-1\right)^2+\left(z+3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y-z\right)^2=0\\\left(y-1\right)^2=0\\\left(z+3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-z=0\\y-1=0\\z+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\\z=-3\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(-2;1;-3\right)\) thỏa mãn yêu cầu bài toán.

61. D
62. C
63. D
64. C
65. C
66. B
67. A
68. D
69. A