a) Ta có: \(\sqrt{27\left(\sqrt{5}-\sqrt{3}\right)^2}\)

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

\(=3\sqrt{15}-9\)

c) Ta có: \(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}=x-\sqrt{x}+1\)

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)

a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)

\(=3-2=1\)

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

b: B<A

=>B-1<0

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

=>-1/căn x<0

=>căn x>0

=>x>0 và x<>1

\(a,\) ta có : 

\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)

\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)

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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :

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

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

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

\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)

vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)

để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

vậy để \(B=2\) thì \(x=4\)

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

$\Leftrightarrow (x-2)(x-3)-2.\frac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow (x-3)[(x-2)-\frac{2}{\sqrt{x-2}+1}]=0$

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

$\Leftrightarrow a^3+a^2-2=0$

$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$

$\Leftrightarrow (a-1)(a^2+2a+2)=0$

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)

Vậy pt có nghiệm duy nhất $x=3$.

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

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

a) Ta có: \(\dfrac{x^2}{y^2}:\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{x^2}{y^2}:\dfrac{x}{y^2}\)

=x

b) Ta có: \(\sqrt{\dfrac{27\left(x-1\right)^2}{12}}+\dfrac{3}{2}-\left(x-2\right)\sqrt{\dfrac{50x^2}{8\left(x-2\right)^2}}\)

\(=\sqrt{\dfrac{9}{4}}\cdot\sqrt{\left(x-1\right)^2}+\dfrac{3}{2}-\left(x-2\right)\cdot\sqrt{\dfrac{25}{4}}\cdot\sqrt{\dfrac{x^2}{\left(x-2\right)^2}}\)

\(=\dfrac{3}{2}\cdot\left(x-1\right)+\dfrac{3}{2}-\left(x-2\right)\cdot\dfrac{5}{2}\cdot\dfrac{x}{2-x}\)

\(=\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}-\dfrac{5}{2}\left(x-2\right)\cdot\dfrac{-x}{x-2}\)

\(=\dfrac{3}{2}x+\dfrac{5}{2}\cdot\left(x\right)\)

=4x

1.\(\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

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

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

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

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

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =-\sqrt{3}\)

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

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

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

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

Câu 3: 

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

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

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

\(=a-2\sqrt{a}\)