a) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{3}.\sqrt{2}+2\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

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

\(=1+\sqrt{2}\)

b) Ta có:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Khi đó:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{961\sqrt{960}+960\sqrt{961}}\)

\(=\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+...+\left(\dfrac{1}{\sqrt{960}}-\dfrac{1}{\sqrt{961}}\right)\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{960}}-\dfrac{1}{\sqrt{961}}\)

\(=1-\dfrac{1}{\sqrt{961}}\)

\(=\dfrac{\sqrt{961}-1}{\sqrt{961}}\)

\(=\dfrac{31-1}{31}\)

\(=\dfrac{30}{31}\)

1. Does your school have a library?

2. How long does it take you to learn English every day?

3. What's your address?

4. We have a thirty-minute break.

5. There are 400 students in Henry's school.

a) \(3.2^4+2^2.3^2-50\)

\(=3.16+4.9-50\)

\(=48+36-50\)

\(=34\)

b) \(4.5^2-3.2^3+3^2.3^2\)

\(=4.25-3.8+9.9\)

\(=100-24+81\)

\(=157\)

c) \(\left(11+159\right).37+\left(185-31\right):14\)

\(=170.37+154:14\)

\(=6290+11\)

\(=6301\)

d) \(3280-\left(3^2.7^3-2^3.49\right)\)

\(=3280-\left(9.343-8.49\right)\)

\(=3280-\left(3087-392\right)\)

\(=3280-3087+392\)

\(=585\)

\(x^2-y^2=1999\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=1999\)

Vì \(1999=1.1999=1999.1=\left(-1\right).\left(-1999\right)=\left(-1999\right).\left(-1\right)\) nên ta có bảng:

Vậy phương trình có nghiệm là: \(\left(x;y\right)=\left(1000;999\right),\left(1000;-999\right),\left(-1000;-999\right)\) và \(\left(-1000;999\right).\)

Câu I.

2. Ta có:

\(\sqrt{a\left(1-b\right)\left(1-c\right)}\)

\(=\sqrt{a\left(1-b-c+bc\right)}\)

\(=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}\) (vì \(a+b+c+2\sqrt{abc}=1\) nên \(a+2\sqrt{abc}=1-b-c\))

\(=\sqrt{a\left(\sqrt{a}+\sqrt{bc}\right)^2}\)

\(=\left|\sqrt{a}+\sqrt{bc}\right|\sqrt{a}\)

\(=\left(\sqrt{a}+\sqrt{bc}\right)\sqrt{a}\) (vì \(\sqrt{a}+\sqrt{bc}>0\))

\(=\sqrt{a^2}+\sqrt{abc}\)

\(=\left|a\right|+\sqrt{abc}\)

\(=a+\sqrt{abc}\) (vì \(a>0\))

Tương tự, ta cũng có:

\(\sqrt{b\left(1-c\right)\left(1-a\right)}=b+\sqrt{abc}\)

\(\sqrt{c\left(1-a\right)\left(1-b\right)}=c+\sqrt{abc}\)

Khi đó:

\(B=\sqrt{a\left(1-b\right)\left(1-c\right)}+\sqrt{b\left(1-c\right)\left(1-a\right)}+\sqrt{c\left(1-a\right)\left(1-b\right)}-\sqrt{abc}\)

\(=\left(a+\sqrt{abc}\right)+\left(b+\sqrt{abc}\right)+\left(c+\sqrt{abc}\right)-\sqrt{abc}\)

\(=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}\)

\(=a+b+c+2\sqrt{abc}\)

\(=1\) (vì \(a+b+c+2\sqrt{abc}=1\))

1. B My friends are drinking coffee.

2. B The girl is crying.

3. C He can't draw.

4. D You musn't be late for school.

5. A We want to visit all the famous sights.

6. B The boys aren't playing badminton.

7. B My father is planting trees in the garden at the moment.

8. C Ms Linda can speak Chinese.

\(25< 5^x< 625.5^2\)

\(5^2< 5^x< 5^4.5^2\)

\(5^2< 5^x< 5^6\)

\(\Rightarrow x\in\left\{3;4;5\right\}.\)

D. 5 số

Sửa đề: \(\sqrt{5x^3+3x^2+3x-2}=\dfrac{x^2}{2}+3x-\dfrac{1}{2}\) \(\left(1\right)\)

Điều kiện: \(5x^3+3x^2+3x-2\ge0\Leftrightarrow x\ge\dfrac{2}{5}\)

\(\left(1\right)\Leftrightarrow2\sqrt{5x^3+3x^2+3x-2}=x^2+6x-1\)

\(\Leftrightarrow2\sqrt{\left(5x^3-2x^2\right)+\left(5x^2-2x\right)+\left(5x-2\right)}=x^2+6x-1\)

\(\Leftrightarrow2\sqrt{x^2\left(5x-2\right)+x\left(5x-2\right)+\left(5x-2\right)}=x^2+x+5x+1-2\)

\(\Leftrightarrow2\sqrt{\left(x^2+x+1\right)\left(5x-2\right)}=\left(x^2+x+1\right)+\left(5x-2\right)\)

\(\Leftrightarrow\left(x^2+x+1\right)-2\sqrt{\left(x^2+x+1\right)\left(5x-2\right)}+\left(5x-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+x+1}-\sqrt{5x-2}\right)^2=0\)

\(\Leftrightarrow\sqrt{x^2+x+1}-\sqrt{5x-2}=0\)

\(\Leftrightarrow\sqrt{x^2+x+1}=\sqrt{5x-2}\)

\(\Leftrightarrow x^2+x+1=5x-2\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

So với điều kiện thấy thỏa mãn.

Vậy phương trình có tập nghiệm là: \(S=\left\{1;3\right\}.\)