a) \(\sqrt{1-x}=\sqrt[3]{8}\) ( ĐK: \(x\le1\) )

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

\(\Leftrightarrow1-x=4\)

\(\Leftrightarrow x=-3\) ( Thỏa mãn )

b) \(\sqrt{4x^2-12x+9}=x+1\) ( ĐK : \(x\ge-1\) )

\(\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+3^2}=x+1\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\\3-2x=x+1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}\) ( Thỏa mãn )

c) \(x+\sqrt{x}-2=0\) ( ĐK : \(x\ge0\) )

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

\(\Leftrightarrow\sqrt{x}-1=0\)

\(\Leftrightarrow x=1\) ( Thỏa mãn )

+) ĐKXĐ : \(x\le1\)

 \(\sqrt{1-x}=\sqrt[3]{8}\)

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

\(\Leftrightarrow1-x=4\)

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

+)  \(\sqrt{4x^2-12x+9}=x+1\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\left(x\ge\frac{3}{2}\right)\\2x-3=-x-1\left(x< \frac{3}{2}\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=3+1\\2x+x=3-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}\left(TM\right)}}\)

+) ĐKXĐ : \(x\ge0\)

 \(x+\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=2\)

+) \(\hept{\begin{cases}\sqrt{x}=1\\\sqrt{x}+1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(TM\right)}\)

+) \(\hept{\begin{cases}\sqrt{x}=2\\\sqrt{x}+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=0\end{cases}}}\left(TM\right)\)

Gọi số cần tìm là ab

Ta có 5 x ab = 3ab

=> 5 x ab = 300 + ab

=> 4 x ab = 300

=> ab = 75

Vậy số cần tim là 375

nhin sai sai kieu gi y so 4 lay o dau ?

thấy mệt

1 : thấy mệt

2 : lưới ( đánh cá )

3 : bóng của mặt trăng ( cơ mà theo vật lý thì nó có thể là bóng của bất cứ vật gì có dạng tròn như vậy )

4 : Môi trường sống

Bài làm:

1) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)

\(=3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)

\(=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)

\(=14\sqrt{3}\)

2) \(\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right).\sqrt{5}+5\sqrt{8}\)

\(=3\sqrt{5}.\sqrt{5}-2\sqrt{10}.\sqrt{5}+\sqrt{5}.\sqrt{5}+5.2\sqrt{2}\)

\(=15-10\sqrt{2}+5+10\sqrt{2}\)

\(=20\)

\(2\sin2x-\cos2x=7\sin x+2\cos x-4\)

\(\Rightarrow4\sin x\cos x-\left(1-2\sin^2x\right)-7\sin x-2\cos x+4=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin^2-7\sin x+3\right)=0\)

\(\Rightarrow2\cos x\left(2\sin x-1\right)+\left(2\sin x-1\right)\left(\sin x-3\right)=0\)

\(\Rightarrow\left(2\sin x-1\right)\left(2\cos x+\sin x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2\sin x-1=0\\2\cos x+\sin x=3,\left(vn\right)\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{\eta}{6}+k2\eta\\x=\frac{5\eta}{6}+k2\eta\end{cases}}}\)

\(\eta=Pi;3.14159\)

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) \(x^8+x^7+1\)

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

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

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

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

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

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

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