1. ĐKXĐ: $x>0; x\neq 9$

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

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)

help

  => đề sai rồi bạn

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

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

2: \(\Leftrightarrow D=\dfrac{4\sqrt{x}+12-x+\sqrt{x}-13}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}=1\)

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

\(\Leftrightarrow-2x\sqrt{x}-x+10x+5\sqrt{x}-2\sqrt{x}-1=x\sqrt{x}+3x+x+3\sqrt{x}+\sqrt{x}+3\)

\(\Leftrightarrow-2x\sqrt{x}+9x-3\sqrt{x}-1=x\sqrt{x}+4x+4\sqrt{x}+3\)

\(\Leftrightarrow-3x\sqrt{x}+5x-7\sqrt{x}-4=0\)

Bạn xem lại đề nhé, nghiệm rất xấu

1.

\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)

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

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

2.

\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)

\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)

ĐKXĐ: x>=0; x<>1

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

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

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

b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)

c: B=2/3

=>căn x/căn x+1=2/3

=>căn x=2

=>x=4

d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)

=>B<1

e: B>1

=>-1/căn x+1>0

=>căn x+1<0(vô lý)

=>KO có x thỏa mãn

f: B nguyên khi căn x chia hết cho căn x+1

=>căn x+1-1 chia hết cho căn x+1

=>căn x+1=1 hoặc căn x+1=-1(loại)

=>căn x=0

=>x=0

ĐKXĐ x\(\ge0,x\ne1,x\ne4\)

(x+2√x+4x√x−8+x+2√x+4x−1):(3+1√x−2+2√x+1)

P=\(\left(\dfrac{\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}+\dfrac{x+2\sqrt{x}+4}{x-1}\right):\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+\sqrt{x}+1+2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

P=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)

P=\(\dfrac{x-1+\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)

P=\(\dfrac{x\sqrt{x}+x-9}{3\left(x-3\right)}\)

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)

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

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{x^2}\)