1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

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

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

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

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

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

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

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

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

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

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

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

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

\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

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

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)

a, ĐKXĐ : \(x\ge1\)

Ta có ; \(PT\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.\sqrt{9}\sqrt{x-1}+24.\sqrt{\dfrac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\dfrac{1}{2}-\dfrac{3}{2}\sqrt{9}+24\sqrt{\dfrac{1}{64}}\right)=-17\)

\(\Leftrightarrow-\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x=290\left(TM\right)\)

Vậy ....

b, ĐKXĐ : \(x\ge3\)

Ta có : \(PT\Leftrightarrow x-3-7\sqrt{x-3}+12=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\) ( TM )

Vậy ..

a) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow-\sqrt{x-1}=-17\)

\(\Leftrightarrow x-1=17^2=289\)

hay x=290

Vậy: S={290}

b) Ta có: \(x-7\sqrt{x-3}+9=0\)

\(\Leftrightarrow x-7\sqrt{x-3}=-9\)

\(\Leftrightarrow x-3-2\cdot\sqrt{x-3}\cdot\dfrac{7}{2}+\dfrac{49}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow\left(\sqrt{x-3}-\dfrac{7}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=4\\\sqrt{x-3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=16\\x-3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=19\\x=12\end{matrix}\right.\)

Vậy: S={19;12}

a \(ĐKXĐ:x\ge1\) 

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3^2}{2}\sqrt{x-1}+\dfrac{24}{8}\sqrt{x-1}=-17\Leftrightarrow\dfrac{1}{2}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\) \(\Leftrightarrow-4\sqrt{x-1}+3\sqrt{x-1}=-17\Leftrightarrow-\sqrt{x-1}=-17\Leftrightarrow\sqrt{x-1}=17\Rightarrow x-1=289\Leftrightarrow x=290\left(TM\right)\) b \(ĐKXĐ:x\ge3\) 

\(\Leftrightarrow x-3-7\sqrt{x-3}+12=0\Leftrightarrow\left(\sqrt{x-3}-3\right)\left(\sqrt{x-3}-4\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=3\\\sqrt{x-3}=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(TM\right)\\x=19\left(TM\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)

=>(x+1)(2x+6)=(2x+1)(x+4)

\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)

=>9x+4=8x+6

=>x=2

b: \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

a: \(\sqrt{5\left(1-a\right)^2}\)

\(=\sqrt{5\left(a-1\right)^2}\)

\(=\sqrt{5}\cdot\sqrt{\left(a-1\right)^2}\)

\(=\sqrt{5}\left|a-1\right|\)

\(=\sqrt{5}\left(a-1\right)\)(do a>1 nên a-1>0)

b: \(\sqrt{\dfrac{9\left|a^2+2a+1\right|}{144}}\)

\(=\sqrt{\dfrac{9}{144}\cdot\left|a^2+2a+1\right|}\)

\(=\sqrt{\dfrac{1}{16}\cdot\left|\left(a+1\right)^2\right|}\)

\(=\sqrt{\dfrac{1}{16}}\cdot\sqrt{\left|\left(a+1\right)^2\right|}\)

\(=\dfrac{1}{4}\cdot\left(a+1\right)^2\)

c: 

ĐKXĐ: x<>5

Sửa đề:\(\dfrac{2}{x-5}\cdot\sqrt{\dfrac{x^2-10x+25}{64}}\)

\(=\dfrac{2}{x-5}\cdot\sqrt{\dfrac{\left(x-5\right)^2}{64}}\)

\(=\dfrac{2}{x-5}\cdot\dfrac{\sqrt{\left(x-5\right)^2}}{\sqrt{64}}\)

\(=\dfrac{2}{x-5}\cdot\dfrac{\left|x-5\right|}{8}\)

\(=\pm\dfrac{1}{4}\)

d: \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{x}-\sqrt{x}\cdot1}{\sqrt{x}-1}\)

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

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

\(a.\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)( x lớn hơn hoặc =1)
\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}\)+2=0
\(\sqrt{x-1}\left(1+\sqrt{4}-\sqrt{25}\right)=-2\)
\(\sqrt{x-1}\left(1+2-5\right)=-2\)
\(\sqrt{x-1}.\left(-2\right)=-2\)
\(\sqrt{x-1}=-2.2\)
\(\sqrt{x-1}-4\)(ko thỏa mãn)
b)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\sqrt{x-1}=-17\)
\(7\sqrt{x-1}=-17\)
\(\sqrt{x-1}=-\dfrac{17}{7}\)(ko thỏa mãn căn bậc 2 ko có số âm)

a: Ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

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

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

\(\Leftrightarrow x-1=1\)

hay x=2

Câu 1 : 

a, \(=8+4-2.6=12-12=0\)

b, đk : x > 0 ; x khác 1 

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

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{5}{4}\)

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

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

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)