a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)

a: ĐKXĐ: x^2-2x<>0 và x^2-1>0

=>(x>1 và x<>2) hoặc x<-1

b: ĐKXĐ: x+1>0 và 5-3x>0

=>x>-1 và 3x<5

=>-1<x<5/3

c: DKXĐ: 5x+3>=0 và 3-x>0

=>x>=-3/5 và x<3

=>-3/5<=x<3

d: ĐKXĐ: 4-x^2>0 và 1+x>=0

=>x^2<4 và x>=-1

=>-2<x<2 và x>=-1

=>-1<=x<2

e: ĐKXĐ: 2-3x<>0 và 1-6x>0

=>x<>2/3 và x<1/6

=>x<1/6

\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)

\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)

Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)

a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)

=>x+6<0

=>x<-6

b: ĐKXĐ: (-2)/(6-x)>=0

=>6-x<0

=>x>6

c: ĐKXĐ: (-x+3)/(-6)>=0

=>-x+3<=0

=>-x<=-3

=>x>=3

d: ĐKXĐ: (7x-1)/-9>=0

=>7x-1<=0

=>x<=1/7

e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0

=>x+2>=0

=>x>=-1

f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0

=>x-2>=0

=>x>=2

`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)

\(a,=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}=-1-3\sqrt{15}\)

\(b,=x\sqrt{2\left(x+1\right)}+\sqrt{\dfrac{2\left(x+1\right)^2}{x+1}}-\sqrt{\dfrac{16\left(x+1\right)}{2}}\\ =x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\\ =\sqrt{2\left(x+1\right)}\left(x+1-2\right)=\left(x-1\right)\sqrt{2\left(x+1\right)}\)

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

\(=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{5-3}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{5-3}\)

=\(\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}\)

=\(-1-3\sqrt{15}\)

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

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

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

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

a.

ĐKXĐ: \(x>0\)

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

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)

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

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)

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

a) Đk: \(x>0;x\ne9;x\ne25\)

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

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

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

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

\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)

b) Đk: \(x\ge0;x\ne1;x\ne25\)

Biểu thức

\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)

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

Đk: \(x>0;x\ne4\)

Biểu thức:

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

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

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

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

Khoảng cách hai ta là 100 bước, em bước 100, anh lùi 1