ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

\(\Leftrightarrow2x-1-2\sqrt{\left(2x-1\right)\left(x+3\right)}-x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}-2\sqrt{x+3}\right)-x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left(\sqrt{2x-1}-x\right)\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=x\left(x\ge0\right)\\\sqrt{2x-1}=2\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x^2\\2x-1=4\left(x+3\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{2}\left(loại\right)\end{matrix}\right.\)

\(3,\\ a,\dfrac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\\ =\dfrac{\sqrt{x}-2\sqrt{x}+1}{1-\sqrt{x}}=\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}=1-\sqrt{2}\)

\(b,\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{1+\sqrt{xy}}\\ =\dfrac{x+2\sqrt{xy}+y}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{1+\sqrt{xy}}\\ =\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1+\sqrt{6}}=\dfrac{5+2\sqrt{6}}{1+\sqrt{6}}\\ =\dfrac{\left(5+2\sqrt{6}\right)\left(\sqrt{6}-1\right)}{5}\\ =\dfrac{3\sqrt{6}+7}{5}\)

Bài 2

b)\(\overrightarrow{AN}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(\overrightarrow{AK}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AN}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)

d)\(S_{ABC}=24\Leftrightarrow\dfrac{1}{2}AN.BC=24\Leftrightarrow AN=6\left(cm\right)\)

\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\left|2.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\right|=\left|2\overrightarrow{AN}\right|=2.AN=12\left(cm\right)\)

Bài 3:

b)\(\overrightarrow{BG}=\overrightarrow{BC}+\overrightarrow{CG}=\overrightarrow{BC}+\dfrac{3}{4}\overrightarrow{CA}=\overrightarrow{BC}+\dfrac{3}{4}\left(\overrightarrow{BA}-\overrightarrow{BC}\right)=\dfrac{1}{4}\overrightarrow{BC}+\dfrac{3}{4}\overrightarrow{BA}=\dfrac{1}{4}\overrightarrow{v}+\dfrac{3}{4}\overrightarrow{u}\)

c)Nhìn hình thấy ko thẳng nên đề sai

Part 1

1.girls

2.are

3.reading

4.boy

5.she's

Part 2

2. No, they aren't. Some girls are making puppets

3. They are red, blue and yellow

4. The girl is reading a text

5. The teacher is in the office

6. Yes, she is reading a book

6.No, they aren't

7.They are painting red, blue and yellow.

8.The girl is reading a text.

9.Our teacher is in the office.

10.Yes, she is.

\(a,5\sqrt{25a^2}-25a\left(a< 0\right)=5\cdot\left|5a\right|-25a=-25a-25a=50a\\ b,\sqrt{49a^2}+3a\left(a>0\right)=\left|7a\right|+3a=10a\\ c,\sqrt{64a^2}-8a\left(a\in R\right)=\left|8a\right|-8a=\left[{}\begin{matrix}0,\forall a\ge0\\-16a,\forall a< 0\end{matrix}\right.\\ d,\sqrt{9a^6}-6a^3=\left|3a^3\right|-6a^3=\left[{}\begin{matrix}-3a^3,\forall a\ge0\\-9a^3,\forall a< 0\end{matrix}\right.\)

gì vậy?

chữ nhỏ quá!

Thôi buồn lắm k làm đâu đi ngủ

1 Those trees are watered by me everyday

2 Many friends were invited to his birthday party by Jack last week

3 Tea is being made in the kitchen by my dad

4 A lot of water should be drunk by us everyday

5 The painting has just been finished by Jane

6 A serious accident was caused in this street by some young boys last night

7 More doctors and nurses will be sent to the northern area by many hospitals next ......

8 Some rules weren't obeyed at school by John and Jack

9 Was your bike stolen by Jackson

10 Vegetables won't be grown by Mr Bill next winter

11 This factory wasn't built by the German during the Second World War

12 This house and these cars are going to be painted for Christmas

13 THe cats had been fed before she went to the cinema

13 Moon cakes were being made by us at this time last week

14 His work had been finished by Jack before 8 am

`#3107.101107`

a)

\(\dfrac{-22}{25}+\left(\dfrac{3}{7}-0,12\right)\\ =\dfrac{-22}{25}+\dfrac{3}{7}-0,12\\ =-\dfrac{79}{175}-0,12\\ =-\dfrac{4}{7}\)

b)

\(\dfrac{3}{8}-\left(1,2-\dfrac{5}{8}\right)\\ =\dfrac{3}{8}-1,2+\dfrac{5}{8}\\ =\dfrac{8}{8}-1,2\\ =1-1,2\\ =-0,2\)

c)

\(\dfrac{4}{15}-\left(2,9-\dfrac{11}{15}\right)\\ =\dfrac{4}{15}-2,9+\dfrac{11}{15}\\=\dfrac{15}{15}-2,9\\ =1-2,9\\ =-1,9\)

d)

\(\dfrac{17}{11}-\left(\dfrac{6}{5}-\dfrac{16}{11}\right)+\dfrac{26}{5}\\ =\dfrac{17}{11}-\dfrac{6}{5}+\dfrac{16}{11}+\dfrac{26}{5}\\ =\left(\dfrac{17}{11}+\dfrac{16}{11}\right)+\left(-\dfrac{6}{5}+\dfrac{26}{5}\right)\\ =\dfrac{33}{11}+\dfrac{20}{5}\\ =3+4\\ =7\)

e)

\(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{7}\right)\\ =\dfrac{39}{5}+\dfrac{9}{4}-\dfrac{9}{5}-\dfrac{5}{4}-\dfrac{6}{7}\\ =\left(\dfrac{39}{5}-\dfrac{9}{5}\right)+\left(\dfrac{9}{4}-\dfrac{5}{4}\right)-\dfrac{6}{7}\\ =\dfrac{30}{5}+\dfrac{4}{4}-\dfrac{6}{7}\\ =6+1-\dfrac{6}{7}\\=\dfrac{43}{7}\)