câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

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

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

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

ĐKXĐ: x \(\ge\)0; x khác 9 (1)

a) B = \(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)

B = \(\frac{-\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{-4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\)

B = \(\frac{4}{3-\sqrt{x}}\)

b) B > A <=> \(\frac{4}{3-\sqrt{x}}>1\) <=> \(\frac{4}{3-\sqrt{x}}-1>0\)

<=> \(\frac{4-3+\sqrt{x}}{3-\sqrt{x}}>0\)

<=> \(\frac{\sqrt{x}+1}{3-\sqrt{x}}>0\)

Do \(\sqrt{x}+1>0\) => \(3-\sqrt{x}>0\) <=> \(\sqrt{x}< 3\)

<=> \(x< 9\)

Kết hợp với đk (1)

=> \(0\le x< 9\)

a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

=>căn x-1>0

=>x>1

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

\(a,A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}.\)

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

\(A=\left(x-3\right)-\left(x+3\right)\)

\(b,\) Ta có : \(A=1=\left(x-3\right)-\left(x+3\right)\)

                                   \(\Leftrightarrow1=x-3-x-3\Leftrightarrow1=-6\left(ko\right)tm\)

Vậy ko có giá trị của x.

mk ko biết đâu

mk mới hok lớp 5 thui

chúc bạn hok tốt nhé

kb với mk nha

Txđ: x thuộc R

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

Th1:  với x>=3 thì A= x-3-x-3= -6

TH2: với x thuộc [-3,3) thì A = -x +3 -x-3= -2x

Th3: với x < -3 thì A = -x+3+x+3 = 6

b. A=1 thuộc TH2 câu a

-2x=1 => x= -1/2 thỏa mãn x thuộc (-3,3) vậy A=1 khi x=-1/2

a: ĐKXĐ: \(x>0\)

b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

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

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