Gấp lắm ah @Như Trương Thị

1. - That student doesn't study hard everyday.

    - Does that student study hard everyday?

2. - Mr brown doesn't go to his office everyday.

    - Does Mr brown go to his office everyday?

3. - I don't want cream and sugar.

    - Do I want cream and sugar?

4. - The children don't go to bed very early.

    - Do the children go to bed very early?

5. - That girl doesn't come from the USA.

    - Does that gir come from the USA?

6. - There isn't a flower vase on the table.

    - Is there a flower vase on the table? 

7. - Lan doesn't wear uniform on weekdays.

    - Does Lan wear uniform on weekdays?

8. - There aren't some students in the library.

    - Are there some students in the library?

Is 189,Tran Quoc Toan street.Group 8,zone 9,Hong Ha ward.

Sorry,i'm use Google Translate.

My address is 3 area , Bac Thanh villag , Yen Thanh distric , Nghe An Province 

 Dịch : Địa chỉ của tôi là : Xóm 3 , Xã Bắc Thành , huyện Yên Thành , tỉnh Nghệ An

Học tốt 

P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)

Ta có : 

\(2a^2+24a+80=2a^2+24a+72+8=2\left(a+6\right)^2+8\)

Vì \(\left(a+6\right)^2\ge0\forall a\Rightarrow2\left(a+6\right)^2+8\ge8\) 

Dấu "=" xảy ra \(\Leftrightarrow2\left(a+6\right)^2=0\Leftrightarrow a+6=0\Leftrightarrow a=-6\)

Vậy GTNN của bt trên là 8 <=> a = - 6

Ta có : 

\(2a^2+24a+80=2a^2+24a+72+8=2\left(a+6\right)^2+8\)

Vì \(\left(a+6\right)^2\ge0\forall a\Rightarrow2\left(a=6\right)^2+8\ge8\)

Dấu '=' xảy ra \(\Leftrightarrow2\left(a+6\right)^2=0\)

\(\Leftrightarrow a+6=0\Leftrightarrow a=-6\)

Vậy GTNN của biểu thức trên là 8 .\(\Leftrightarrow a=-6\)

Bài 2 :                                                               Bài giải

\(a,\text{ }\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3\)

\(=\frac{9}{10}-\frac{7}{10}+9,3=\frac{1}{5}+9,3=0,2+9,3=9,5\)

\(b,\text{ }\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)

\(c,\text{ }\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)

Bài 2 : 

a ) \(\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3\)

\(=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3=\frac{9}{10}-\frac{7}{10}+9,3\)

\(=\frac{1}{5}+9,3=0,2+9,3=9,5\)

b ) \(\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}\)

\(=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)

c ) \(\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}\)

\(=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)

c) Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9

A = \(-\frac{1}{\sqrt{x}-3}\) => -2A = \(\frac{2}{\sqrt{x}-3}\)

Để -2A thuộc Z <=> \(2⋮\sqrt{x}-3\)

<=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lập bảng: 

Vậy ....

1. Thay x = 1 vào đa thức f (x) = ax² + bx + c . Ta có :

f ( x ) = a.1² + b.1 + c

         = a + b + c

         = 0

Vậy x = 1 là nghiệm của f ( x )

Bài 1 :

Giả sử x = 1 là nghiệm của đa thức f (x) = ax² + bx + c

=> f (x) = a . 1² + b . 1 + c = 0

<=> f(x) = a + b + c = 0 

Vậy nếu a + b + c = 0 thì x = 1 là nghiệm của đa thứ f (x)

Bài 2 :

a) \(\left(x-2\right)\left(2x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

Vậy nghiệm của đa thức là x=2 hoặc x=4

b) \(\left(3x-9\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-9=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

Vậy .................

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

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

\(\Leftrightarrow x=3\)

Vậy .............

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

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

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

Vậy...............

a. \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Rightarrow10x^2-35x+16x-10x^2=5\)

\(\Rightarrow-19x=5\)

\(\Rightarrow x=-\frac{5}{19}\)

b. \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

\(\Rightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)

\(\Rightarrow\frac{7}{6}x=\frac{1}{4}\)

\(\Rightarrow x=\frac{3}{14}\)

c. \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Rightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Rightarrow-14x+5=x-2\)

\(\Rightarrow-14x-x=-2-5\)

\(\Rightarrow-15x=-7\)

\(\Rightarrow x=\frac{7}{15}\)

a, \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Leftrightarrow10x^2-35x+16x-10x^2=5\)

\(\Leftrightarrow-19x=5\Leftrightarrow x=-\frac{5}{19}\)

b, \(x\left(x-\frac{1}{3}\right)-\frac{1}{2}x\left(2x-3\right)=\frac{1}{4}\)

\(\Leftrightarrow x^2-\frac{1}{3}x-x^2+\frac{3}{2}x=\frac{1}{4}\)

\(\Leftrightarrow\frac{7}{6}x=\frac{1}{4}\Leftrightarrow x=\frac{3}{14}\)

c, \(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)

\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Leftrightarrow-15x+7=0\Leftrightarrow x=\frac{7}{15}\)