Trc khi tớ lm , mong câu đọc cho : I'm very lazy, I'll only do 3 sentences, if you need me to finish it !!! 

1, Keeping the environment clean is very important.

=> It's very important to keep the enviroment clean

2,"Do you have to do your homework tonight". She said to me

=> She asked me if I had to do my homework that night

3, "Why don't we go out for a walk"

=> My father suggested going out for a walk

ns tớ sai đâu sửa nha 

ok, thank Brainchild

đẶT \(A=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)

\(=\sqrt{\frac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{11}}-\sqrt{\frac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{13}}\)

\(=\sqrt{\frac{18-3\sqrt{3}-8\sqrt{3}+4}{11}}-\sqrt{\frac{5\sqrt{3}+6+20+8\sqrt{3}}{13}}\)

\(=\sqrt{\frac{11\left(2-\sqrt{3}\right)}{11}}-\sqrt{\frac{13\left(2+\sqrt{3}\right)}{13}}\)

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

ta có: \(2-\sqrt{3}< 2+\sqrt{3}\Rightarrow\sqrt{2-\sqrt{3}}< \sqrt{2+\sqrt{3}}\)

\(\Rightarrow A< 0\Rightarrow-A>0\)

\(\Rightarrow-A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

\(A^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2\)

\(A^2=\left(\sqrt{2+\sqrt{3}}\right)^2-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\left(\sqrt{2-\sqrt{3}}\right)^2\)

\(A^2=\left|2+\sqrt{3}\right|-2\sqrt{4-3}+\left|2-\sqrt{3}\right|\)

\(A^2=2+\sqrt{3}-2+2-\sqrt{3}\)

\(A^2=2\)

\(A=\pm\sqrt{2}\)

mà -A > 0 nên A = \(-\sqrt{2}\)

~~ Học tốt ~~

Ở dòng: 

\(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\) còn có thêm cách phân tích

\(\sqrt{2}.A=\sqrt{4-2.\sqrt{3}}-\sqrt{4+2.\sqrt{3}}\)

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

\(=\sqrt{3}-1-\sqrt{3}-1=-2\)

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

\(ĐKXĐ:x\ge\frac{-1}{3}\)

Bình phương hai vế:

\(\left(\sqrt{5x+7}-\sqrt{x+3}\right)^2=\left(\sqrt{3x+1}\right)^2\)

\(\Leftrightarrow6x+10-2\sqrt{\left(5x+7\right)\left(x+3\right)}=3x+1\)

\(\Leftrightarrow3x+9-2\sqrt{\left(5x+7\right)\left(x+3\right)}=0\)

\(\Leftrightarrow3\left(x+3\right)-2\sqrt{\left(5x+7\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(3\sqrt{x+3}-2\sqrt{5x+7}\right)=0\)

+) \(\sqrt{x+3}=0\Rightarrow x=-3\left(ktmđk\right)\)

+) \(3\sqrt{x+3}=2\sqrt{5x+7}\)

Bình phương hai vế: \(9\left(x+3\right)=4\left(5x+7\right)\)

\(\Leftrightarrow9x+27=20x+28\Leftrightarrow-11x=1\Leftrightarrow x=\frac{-1}{11}\)(ktm)

Vậy phương trình vô nghiệm

Nghiệm \(\frac{-1}{11}\)thỏa mãn nha, mk nhầm

https://englishshowroom.webs.com/DIRECT%20INTO%20INDIRECT%20SPEECH%20Answers.pdf

Đề có nhầm k vậy????

.

.

Đề ko nhầm nha!!! giải giúp mk đi

ĐK: \(x\ge0\)

+) Với x = 0 => A = 0

+) Với x khác 0

Ta có: \(\frac{1}{A}=\frac{3}{4}\sqrt{x}-\frac{3}{4}+\frac{3}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)-\frac{3}{4}\ge\frac{3}{4}.2-\frac{3}{4}=\frac{3}{4}\)

=> \(A\le\frac{4}{3}\)

Dấu "=" xảy ra <=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\)<=> x = 1

Vậy max A = 4/3 tại x = 1

Còn có 1 cách em quy đồng hai vế giải đenta theo A thì sẽ tìm đc cả GTNN và GTLN 

\(A=\frac{1}{9}\left(3x-2y-2\right)^2+\frac{5}{9}\left(y-2\right)^2+8\ge8\)

Can downvote-er this answer explain us, why you do that? If you don't understand something, I ready to help you!