Ta có: \(a^3-a=a\left(a^2-1\right)=\left(a-1\right)a\left(a+1\right)\)

Vì \(a\left(a-1\right)\) là tích của hai số tự nhiên liên tiếp=> \(a\left(a-1\right)⋮2\Rightarrow a^3-a⋮2\left(1\right)\)

Mặt khác: \(a\left(a-1\right)\left(a+1\right)\) là tích của 3 số tự nhiên liên tiếp nên trong 3 số có 1 số chia hết cho 3=> \(a^3-a⋮3\left(2\right)\)

Từ (1) (2) kết hợp với \(\left(2,3\right)=1\Rightarrow a^3-a⋮2.3\Leftrightarrow a^3-a⋮6\)

Tương tự: \(b^3-b⋮6,c^3-c⋮6\)

\(\Rightarrow\left(a^3-a\right)+\left(b^3-b\right)+\left(c^3-c\right)=a^3+b^3+c^3-a-b-c⋮6\left(đpcm\right)\)

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

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt[]{x}+1\right)}\right)\)

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

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

Ta có:

\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\left(1\right)\)

\(\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\Leftrightarrow\dfrac{2x}{40}=\dfrac{y}{10}=\dfrac{4z}{60}=\dfrac{2x-y+4z}{40-10+60}=\dfrac{270}{90}=3\)

\(\Rightarrow\dfrac{x}{20}=3\Rightarrow x=60\)

\(\Rightarrow\dfrac{y}{10}=3\Rightarrow y=30\)

\(\Rightarrow\dfrac{z}{15}=3\Rightarrow z=45\)

sai r a ạ 

\(\dfrac{3^5.3^7}{3^{11}}=\dfrac{3^{12}}{3^{11}}=3\)

a) \(D=4\sqrt{\dfrac{1}{3}}+5\sqrt{12}-6\sqrt{27}\)

\(=\dfrac{4}{9}\sqrt{3}+5.2\sqrt{3}-6.3\sqrt{3}\)

\(=\dfrac{4}{9}\sqrt{3}+10\sqrt{3}-18\sqrt{3}\)

\(=-\dfrac{68}{9}\sqrt{3}\)

b) \(E=\dfrac{2}{\sqrt{3}-1}-\sqrt{4-2\sqrt{3}}\)

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

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

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

\(=\sqrt{3}+1-\sqrt{3}+1=2\)

c) \(F=\dfrac{\sqrt{15}-\sqrt{10}}{\sqrt{3}-\sqrt{2}}+\dfrac{3}{2-\sqrt{5}}\)

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

\(=\sqrt{5}-6-3\sqrt{5}=-2\sqrt{5}-6\)

1. abroad

2. country

3. difficulties

4. fluent

5. achieved

sao câu 9 lại là a v a

1. A

2. D

3. D

4. B

5. A

6. C

7. B

8. C

9. C

10. B

\(\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-3\sqrt{8^2\left(y-2\right)}+4\sqrt{7^2\left(y-2\right)}\)

\(=\sqrt{y-2}-3.8\sqrt{y-2}+4.7\sqrt{y-2}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}\)

\(=5\sqrt{y-2}\)