1. He always ____travels______________ to work by motoribike.

2. What time_will he_______________here?- He ___will be_________ here in half an hour.

3. The sea _____covers____________most of the surface of the earth.

4. When the Queen____arrives__________, the crowd will stand up.

5. I___will be_______________ready in five minutes.

6. I want to speak to Professor Wilson when he____is______free?

7. He____ comes_____before you leave.

8. Many people ____will attend______________ the concert tomorrow night.

9. _____Does he help__________us when he is free?

10. My father ____works______________very hard ever day.

Ta có: \(x^{10}=25\cdot x^8\)

- Với x=0 => x thỏa man đề bài.

- Với x khác 0 ta có: 

\(x^{10}=25\cdot x^8\)

<=> \(\frac{x^{10}}{x^8}=25\)

<=> \(x^2=25\)

<=> \(x=\pm5\)

Vậy x\(\in\left\{-5;0;5\right\}\)

a) Đề bài phải là : \(\left(x+y\right)^2-\left(x-y\right)^2\)thì mới phân tích được.

Nếu đề bài như trên ta có:

 \(\left(x+y\right)^2-\left(x-y\right)^2=\)\(\left(x+y-x+y\right)\left(x+y+x-y\right)=2x\cdot2y=4xy\)

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

            = \(2x\cdot\left(4x+2\right)=2x\cdot2\cdot\left(2x+1\right)=4x\cdot\left(2x+1\right)\)

c) Ta có : \(x^3+y^3+z^3-3xyz\)

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

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

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

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

Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)

Ta có: C(x) =\(x^2-9x+20=x^2-4x-5x+20=\left(x-4\right)\left(x-5\right)\)

Vậy nghiệm của C(x) là x\(\in\left\{4;5\right\}\)

Ta có: D(x)\(=4x^2+4x+1=\left(2x+1\right)^2\)

Vậy D(x) có nghiệm x=-1/2

Ta có: E(x)=\(2\left(x-1\right)-5\left(x-2\right)=2x-2-5x +10\)= \(8-3x\)

Vậy E(x) có nghiệm x=8/3

Ta có: F(x)=\(2x^2-5x+2=\left(2x^2-x\right)-\left(4x-2\right)\)= \(\left(x-2\right)\left(2x-1\right)\)

Vậy F(x) có nghiệm là x\(\in\left\{\frac{1}{2};2\right\}\)

a) ĐKXĐ: \(x\ne\left\{-3;-\frac{1}{3}\right\}\)

Ta có: \(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=\)\(\frac{\left(3x-1\right)\left(x+3\right)+\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}\)=\(\frac{3x^2+9x-x-3+3x^2+x-9x-3}{3x^2+9x+x+3}\)

          =  \(\frac{6x^2-6}{3x^2+10x+3}\)

=>  \(\frac{6x^2-6}{3x^2+10x+3}=2\)

<=> \(6x^2-6=6x^2+20x+6\)

<=> 20x=12

<=>x=\(\frac{12}{20}=\frac{3}{5}\)

Vậy x=3/5

a) take

b) does....do

c) will enjoy

d) Do...do

e) Will....play

nửa giờ