a) đk: \(x\ge-2\)

Ta có: \(\sqrt{x+2}-\sqrt{4x+8}+\frac{3}{4}\sqrt{9x+18}=3\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}+\frac{9}{4}\sqrt{x+2}=3\)

\(\Leftrightarrow\frac{5}{4}\sqrt{x+2}=3\)

\(\Leftrightarrow\sqrt{x+2}=\frac{12}{5}\)

\(\Leftrightarrow x+2=\frac{144}{25}\)

\(\Rightarrow x=\frac{94}{25}\) (tm)

b) đk: \(x\ge\frac{3}{2}\)

Ta có: \(\sqrt{x^2-4x+4}=2x-3\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=2x-3\\x-2=3-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=\frac{5}{3}\left(tm\right)\end{cases}}\)

a) \(\sqrt{x+2}-\sqrt{4x+8}+\frac{3}{4}\sqrt{9x+18}=3\)

ĐKXĐ : x ≥ -2

⇔ \(\sqrt{x+2}-\sqrt{2^2\left(x+2\right)}+\frac{3}{4}\sqrt{3^2\left(x+2\right)}=3\)

⇔ \(\sqrt{x+2}-2\sqrt{x+2}+\frac{3}{4}\cdot3\sqrt{x+2}=3\)

⇔ \(-\sqrt{x+2}+\frac{9}{4}\sqrt{x+2}=3\)

⇔ \(\frac{5}{4}\sqrt{x+2}=3\)

⇔ \(\sqrt{x+2}=\frac{12}{5}\)

⇔ \(x+2=\frac{144}{25}\)

⇔ \(x=\frac{94}{25}\left(tmđk\right)\)

b) \(\sqrt{x^2-4x+4}=2x-3\)

⇔ \(\sqrt{\left(x-2\right)^2}=2x-3\)

⇔ \(\left|x-2\right|=2x-3\)(1)

Với x < 2

(1) ⇔ -( x - 2 ) = 2x - 3

     ⇔ 2 - x = 2x - 3

     ⇔ -x - 2x = -3 - 2

     ⇔ -3x = -5

     ⇔ x = 5/3 ( tm )

Với x ≥ 2

(1) ⇔ x - 2 = 2x - 3

     ⇔ x - 2x = -3 + 2

     ⇔ -x = -1

     ⇔ x = 1 ( ktm )

Vậy x = 5/3

Phương pháp tách khá dễ thôi

Ta có: \(\sqrt{43-30\sqrt{2}}\)

\(=\sqrt{25-30\sqrt{2}+18}\)

\(=\sqrt{\left(5\right)^2-2\cdot5\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2}\)

\(=\sqrt{\left(5-3\sqrt{2}\right)^2}\)

\(=\left|5-3\sqrt{2}\right|\)

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

Đặt x/-4=k => x=-4k

      y/-7=k => y=-7k

      z/3=k  => z=3k

=> A=8k+7k+15k / -8k+21k-18k

     A=30k / -5k

=> A=-6 

1, The doctor told Elena that she should sleep more or she try to relax more.

2, Eating carrots will help you see object more clearly at night.

3,  Vitamins play an important role on our diet

The flat is big but our old one was bigger.

-> The flag is smaller than the old one.

-> The flag isn't as big as the old one.

ĐKXĐ: x \(\ge\)0

Ta có: \(\frac{3\sqrt{x}}{\sqrt{x}+2}=3-\frac{6}{\sqrt{x}+2}\ge3-\frac{6}{2}=0\)

Giá trị nhỏ nhất của \(\frac{3\sqrt{x}}{\sqrt{x}+2}\)bằng 0 tại x = 0

Bài của cô Chi làm hơi tắt =))

Cho \(A=\frac{3\sqrt{x}}{\sqrt{x}+2}=\frac{3\sqrt{x}+6-6}{\sqrt{x}+2}=\frac{3\left(\sqrt{x}+2\right)-6}{\sqrt{x}+2}=3-\frac{6}{\sqrt{x}+2}\ge3-3=0\)

Dấu bằng xảy ra 

\(\Leftrightarrow\frac{6}{\sqrt{x}+2}=3\)

\(\Leftrightarrow\sqrt{x}+2=2\)

\(\Leftrightarrow x=0\)

Vậy GTNN của A = 0 \(\Leftrightarrow x=0\)