a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}-1}{3-\sqrt{a}}\)

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

\(=\dfrac{2\sqrt{a}-9-a+9+2a-5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-3\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}-1}{\sqrt{a}-3}\)

b: A là số nguyên

=>\(\sqrt{a}-3+2⋮\sqrt{a}-3\)

=>\(\sqrt{a}-3\in\left\{1;-1;2;-2\right\}\)

=>a thuộc {16;25;1}

mình ghi thiếu b , căn x + 9 <= 31 

so sánh : -3 căn3  và -2 căn7

\(\sqrt{-2x+1}>7\)

<=> -2x + 1 > 7²

<=> -2x > 49 - 1

<=> -2x > 48

<=> -2x : (-2) > 48 : (-2)

<=> x < -24

\(a) Đk:x<\dfrac{1}{2}\)

\(\sqrt{-2x+1}>7\)

\(\Leftrightarrow\)\(-2x+1>49\)

\(\Leftrightarrow\)\(x<-24\)

\(b)\)\(Đk:x>-9\)

\(\sqrt{x+9}\)\(\le\)\(31\)

\(\Leftrightarrow\)\(x+9\)\(\le\)\(961\)

\(\Leftrightarrow\)\(x\)\(\le\)\(952\)

\(c)\)Ta có:

\(-3\sqrt{3}=-\sqrt{27} \)

\(-2\sqrt{7}=-\sqrt{28}\)

\(-\sqrt{27}>-\sqrt{28}\)

\(\Rightarrow\)\(-3\sqrt{3}>-2\sqrt{7}\)

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

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

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

b: Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1-2⋮\sqrt{x}+1\)

=>căn x+1 thuộc {1;2}

=>căn x thuộc {0;1}

mà x<>1

nên x=0

ghi đề rõ ra đi bạn ơi. ghi vậy sao hiểu đc

Đề n chỉ ghi là giải pt thôi bạn

a) ĐK:  \(x\ge0\)

  \(x-12\sqrt{x}+32=0\)

\(\Leftrightarrow\)\(x-4\sqrt{x}-8\sqrt{x}+32=0\)

\(\Leftrightarrow\)\(\left(\sqrt{x}-4\right)\left(\sqrt{x}-8\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x}-4=0\\\sqrt{x}-8=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=16\\x=64\end{cases}}\) (t.m)

Vậy...

b) ĐK:  \(x\ge-\frac{1}{2}\)

 \(\sqrt{x^2-x+9}=2x+1\)

\(\Leftrightarrow\)\(x^2-x+9=4x^2+4x+1\)

\(\Leftrightarrow\)\(3x^2+5x-8=0\)

\(\Leftrightarrow\)\(3x^2-3x+8x-8\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(3x+8\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\left(L\right)\end{cases}}\)

Vậy...

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

\(x< 0\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x< 0\end{matrix}\right.\Leftrightarrow x\left(x-1\right)>0\Rightarrow\left|x\left(x-1\right)\right|=x\left(x-1\right)=x^2-x\)

\(b,\sqrt{13x}.\sqrt{\frac{52}{x}}=\sqrt{\frac{13.52.x}{x}}=\sqrt{13.52}=\sqrt{13^2.2^2}=\sqrt{26^2}=26\)

Lời giải :

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

b) \(\sqrt{13x}\cdot\sqrt{\frac{52}{x}}=\sqrt{\frac{13x\cdot52}{x}}=\sqrt{676}=26\)

c) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)

d) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)

ĐKXĐ : \(-1\le x\le9\)

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

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

\(\Leftrightarrow\left(x-6\right)^2=4\left(9-x\right)\)

\(\Leftrightarrow x^2-12x+36=36-4x\)

\(\Leftrightarrow x\left(x-8\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=8\end{array}\right.\)

Thử lại được x = 8 thỏa mãn pt.