`a)sqrt{x^2-2x+1}=2`

`<=>sqrt{(x-1)^2}=2`

`<=>|x-1|=2`

`**x-1=2<=>x=3`

`**x-1=-1<=>x=-1`.

Vậy `S={3,-1}`

`b)sqrt{x^2-1}=x`

Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)

`<=>x>=1`

`pt<=>x^2-1=x^2`

`<=>-1=0` vô lý

Vậy pt vô nghiệm

`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`

`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`

`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`

`<=>2sqrt{x-5}=4`

`<=>sqrt{x-5}=2`

`<=>x-5=4`

`<=>x=9(tmđk)`

Vậy `S={9}.`

`d)x-5sqrt{x-2}=-2(x>=2)`

`<=>x-2-5sqrt{x-2}+4=0`

Đặt `a=sqrt{x-2}`

`pt<=>a^2-5a+4=0`

`<=>a_1=1,a_2=4`

`<=>sqrt{x-2}=1,sqrt{x-2}=4`

`<=>x_1=3,x_2=18`,

`e)2x-3sqrt{2x-1}-5=0`

`<=>2x-1-3sqrt{2x-1}-4=0`

Đặt `a=sqrt{2x-1}(a>=0)`

`pt<=>a^2-3a-4=0`

`a-b+c=0`

`<=>a_1=-1(l),a_2=4(tm)`

`<=>sqrt{2x-1}=4`

`<=>2x-1=16`

`<=>x=17/2(tm)`

Vậy `S={17/2}`

d.

ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:

$a^2+2-5a=-2$

$\Leftrightarrow a^2-5a+4=0$

$\Leftrightarrow (a-1)(a-4)=0$

$\Rightarrow a=1$ hoặc $a=4$

$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$

$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)

e. ĐKXĐ: $x\geq \frac{1}{2}$

Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:

$a^2+1-3a-5=0$

$\Leftrightarrow a^2-3a-4=0$

$\Leftrightarrow (a+1)(a-4)=0$

Vì $a\geq 0$ nên $a=4$

$\Leftrightarrow \sqrt{2x-1}=4$

$\Leftrightarrow x=\frac{17}{2}$

a.

$\sqrt{x^2-2x+1}=2$

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

$\Leftrightarrow |x-1|=2$

$\Rightarrow x-1=\pm 2$

$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)

b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$

PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)

Vậy pt vô nghiệm

c. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

a. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 1=2\sqrt{x-2}$

$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)

a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)

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

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

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

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

Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)

c: C=-1

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

=>\(\sqrt{x}=2\)

=>x=4(nhận)

d: C>0

=>\(2\sqrt{x}-3< 0\)

=>\(\sqrt{x}< \dfrac{3}{2}\)

=>\(0< =x< \dfrac{9}{4}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)

giups mình với

b: Ta có: \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

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

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

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

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

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

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

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

\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

\(f,\sqrt{8-2\sqrt{15}}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)

\(\Rightarrow2\sqrt{5}-\sqrt{3}+\sqrt{3}\)

\(g,\sqrt{9-2\sqrt{4}}-\sqrt{9+2\sqrt{14}}\)

\(\Leftrightarrow\sqrt{9-2\times2}-\sqrt{\left(\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow\sqrt{9-4}-\left(\sqrt{2}+\sqrt{7}\right)\)

\(\Rightarrow\sqrt{5}-\sqrt{2}-\sqrt{7}\)

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)

\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)

\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)

\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)

\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)

\(\Leftrightarrow\) Pt vô nghiệm

Vậy không có giá trị x thỏa yêu cầu đề bài.

a. Áp dụng công thức L'Hospital:

\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt[3]{x+1}-\sqrt{1-x}}=\lim\limits_{x\to 0}\frac{\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}{\frac{1}{3}(x+1)^{\frac{-2}{3}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}\)

b.

\(\lim\limits_{x\to 0}(\frac{1}{x}-\frac{1}{x^2})=\lim\limits_{x\to 0}\frac{x-1}{x^2}=-\infty\)

c. Áp dụng quy tắc L'Hospital:

\(\lim\limits_{x\to +\infty}\frac{x^4-x^3+11}{2x-7}=\lim\limits_{x\to +\infty}\frac{4x^3-3x^2}{2}=+\infty \)

d.

\(\lim\limits_{x\to 5}\frac{7}{(x-1)^2}.\frac{2x+1}{2x-3}=\frac{7}{(5-1)^2}.\frac{2.5+11}{2.5-3}=\frac{11}{16}\)