1. This is the first time Khanh went to Japan .

=> Khanh hasn't gone to Japan before .

2. Uyen stared learning 2 weeks ago .

=> Uyen has learnt for 2 weeks .

3. My parents began drinking when it started to rain .

=> My parents have drunk since it started to rain .

4. Tung last had his car repaired when I left him .

=> Tung haven't had his car repaired since I left him.

5. When did she have it ?

=> How long have she had it ?

6. I haven't seen my grandfather for 2 months .

=> The last time I saw my grandfather was 2 months ago .

7. Gin hasn't taken a bath since Tuesday . 

=> It is Tuesday since Gin took a bath .

Học tốt

1) Khanh hasn't gone to Japan before

2) Uyen has learnt English for 2 weeks

3) My parents have drunk since it started rain

4) Tung hasn't repaired his car since the last time I left him

5) How long is it since she have had it ?

6) The last time I saw my grandfather was 2 months ago

7) It is 6 days since the last time Gin took a bath

Các bạn ơi phần này có 30 người trả lời thì mình sẽ ra phần 2 nhé

1. Là Thánh Gióng.

2.Là Ngô Quyền hay (Ngô Vương)

a) 25/100 ; 12/100 và -35/100

b) -12/32 ; -10/32 và -11/32

Mk hơi mới nên không rõ cách viết

1)

a)\(\frac{25}{100};\frac{12}{100};\frac{-35}{100}\)b)\(\frac{-12}{32};\frac{-10}{32};\frac{-11}{32}\)

2)

a)\(\frac{-1}{2}\&\frac{-4}{5}\)                              b)\(\frac{-6}{7}\&\frac{-7}{8}\)                   c)\(\frac{17}{200}\&\frac{17}{314}\)    

\(\Rightarrow=\frac{-5}{10}\&\frac{-8}{10}\)            \(\Rightarrow=\frac{-48}{56}\&\frac{-49}{56}\)             Mà\(200< 314\)

Mà\(\frac{-5}{10}>\frac{-8}{10}\)                     Mà\(\frac{-48}{56}>\frac{-49}{56}\)           \(\Rightarrow\frac{17}{200}>\frac{17}{314}\)

\(\Rightarrow\frac{-1}{2}>\frac{-4}{5}\)                       \(\Rightarrow\frac{-6}{7}>\frac{-7}{8}\)

a) \(x^5-x^4-1\)

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

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

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

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

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

b) \(x^8+x^7+1\)

\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

Tính

\(E=1-2+3-4+5-6+...+71-72\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(71-72\right)\)  (có 36 cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

\(=\left(-1\right).36=-36\)

Vậy \(E=-36\).

Ta có: E=1-2+3-4+5-6+...+71-72

=> E=(1-2)+(3-4)+(5-6)+...+(71-72)

=> E= (-1)+(-1)+(-1)+...+(-1)

Dãy trên có số sô hạng là: (72-1):1+1=72 (số hạng)

Có số cặp là: 72:2=36(cặp)

=> E=(-1) x 36=-36

Áp dụng phương pháp hệ số bất định để phân tích \(x^4-2x^3-x^2-2x+1\)thành nhân tử.

Phân tích được là: \(\left(x^2-3x+1\right)\left(x^2+x+1\right)\)

=> \(\left(x^2-3x+1\right)\left(x^2+x+1\right)=0\)

Vì \(\left(x^2+x+1\right)>0\Rightarrow x^2-3x+1=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}}{2}+\frac{3}{2}\\x=\frac{-\sqrt{5}}{2}+\frac{3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+3}{2}\\x=\frac{-\sqrt{5}+3}{2}\end{cases}}}\)

\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)

Ta có: \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow\text{ax}=\sqrt{\frac{2a-b}{b}}\Rightarrow1+\text{ax}=1+\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}}\)

\(1-\text{ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)

\(\Rightarrow\frac{1+\text{ax}}{1-\text{ax}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}-\sqrt{2a-b}}=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2b-2a}\left(1\right)\)

 \(bx=\frac{b}{a}\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}\left(2a-b\right)}{a}\Rightarrow\hept{\begin{cases}1-bx=\frac{a-\sqrt{b\left(2a-b\right)}}{a}\\1+bx=\frac{a+\sqrt{b\left(2a-b\right)}}{a}\end{cases}}\)

\(\Rightarrow\frac{1-bx}{1+bx}=\frac{a-\sqrt{b\left(2a-b\right)}}{a+\sqrt{b\left(2a-b\right)}}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{a^2-2ab+b^2}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{\left(a-b\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)

\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)

PLEASE HELP ME!

Bài 1 đề nghị sửa lại : tổ III ủng hộ ít hơn tổng số tiền tổ I và tổ II ủng hộ là 72000đ

Ta có : x² - 2x + 10 = 0

=> x² - 2x + 1 = -9

=> (x - 1)² = -9

=> \(x\in\varnothing\)

\(x^2-2x+10=0\)

\(\Leftrightarrow x^2-2x+1+9=0\)

\(\Leftrightarrow\left(x-1\right)^2+9=0\)

Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\9>0\end{cases}}\)

=> Phương trình vô nghiệm