Đặt x=a-2,ta có : \(P=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

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

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

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

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

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)

\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)

còn lại cậu tự tính nha

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)

\(\frac{11}{6}< x< \frac{32}{9}\)

1,

\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)

\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)

\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)

\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)

\(=\frac{2\sqrt{h-1}}{h-2}\)

Thay \(h=3\)vào D ta có:

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

Vậy với \(h=3\)thì \(D=2\sqrt{2}\)

2,

a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)

Vậy PT có nghiệm là \(x=2\)

b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))

\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)

\(\Leftrightarrow0=-3\)(vô lí)

Vậy PT đã cho vô nghiệm.

Bài 2 :

b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)

ĐKXĐ : \(x\ge1\)

Pt(1) tương đương :

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

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

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)

Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)

\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)

Khi đó pt (*) trở thành :

\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)

\(\Leftrightarrow2\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\) ( Thỏa mãn )

Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)

Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :

\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)

\(\Leftrightarrow2=2\) ( Luôn đúng )

Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)

Bài 1 : 

a) ĐKXĐ : \(-1\le a\le1\)

Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)

\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)

\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)

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

Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)

b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :

\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)

Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)

mn ơi giúp em vs ạ !!!

giúp e vs

a , \(x=2\left(\sqrt{x-1}-\sqrt{x-2}\right)\)

suy ra x =2 

b, x=3 

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