a: \(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: Để A<0 thì căn x-2<0

=>0<x<4

a: \(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: Để A<0 thì căn x-2<0

=>0<x<4

2. He asked her where she was going.

3. He asked which way they had gone.

4. I told her to bring it back if it didn't fit.

5. She told us not to try to open it then.

6. I asked her if it was going to be a fine day that day.

7. She said that he was not at home.

8. The girl wanted to know if the bus station was far away.

9. Tom told Ann not to stay out late.

10. He asked her to let him borrow her car.

11. Thomas asked Jean if he had seen his gloves.

12. I told Mary not to leave the window open.

13. She said that she would have a cup of tea with me.

14. She said that she would pay him if she could.

15. She asked what I was going to do the next summer.

thì vừa là đường trung tuyến vừa là đường trung trực thì tam giác đó cân chứ sao trời!

vẽ hình sẽ ra ngay thôi

Cần chứng minh rõ ràng ạ -.-

​

C

\(ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2}=7-2x\\ \Leftrightarrow\left|x+\dfrac{1}{2}\right|=7-2x\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=7-2x,\forall x+\dfrac{1}{2}\ge0\\x+\dfrac{1}{2}=2x-7,\forall x+\dfrac{1}{2}< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{6},\forall x\ge-\dfrac{1}{2}\left(tm\right)\\x=\dfrac{15}{2},\forall x< -\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{13}{6}\)

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

\(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)

\(=\sqrt{5}+2-\sqrt{5}+2=4\)

\(y=\sqrt{3+2\sqrt{5}}-\sqrt{3-2\sqrt{5}}\)

Xem lại đề, \(\sqrt{3-2\sqrt{5}}\) không xác định.

\(a,\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\\ =\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =2+\sqrt{5}-\sqrt{5}+2=4\)

Bài 5: 

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-20}{1}=-20\)

Do đó: x=-40; y=-60; z=-80

\(\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\\frac{3}{4}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{3}{8}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{3}{8}\right\}\)

Làm đầy đủ hộ mình đi được k bạn

\(a,1\frac{1}{3}+\frac{2}{3}x=5\frac{1}{3}\)

\(\Rightarrow\frac{4}{3}+\frac{2}{3}x=\frac{16}{3}\)

\(\Rightarrow\frac{2}{3}x=\frac{16}{3}-\frac{4}{3}\)

\(\Rightarrow\frac{2}{3}x=4\)

\(\Rightarrow x=4:\frac{2}{3}=4.\frac{3}{2}=6\)

Vậy x=6

\(b,\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\\frac{3}{4}-2x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\-2x=\frac{3}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{-3}{8}\end{cases}}\)

Vậy.................

\(c,\frac{-4}{5}x+1\frac{1}{3}x=-1\frac{1}{3}\)

\(\Rightarrow\frac{-4}{5}x+\frac{4}{3}x=\frac{-4}{3}\)

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

\(\Rightarrow x=\frac{-4}{3}:\frac{-2}{3}=\frac{-4}{3}.\frac{3}{-2}=2\)

Vậy x=2