K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

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

b) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}+1-6\sqrt{x}=0\)

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

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

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{25}\)(nhận)

Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=\dfrac{1}{25}\)

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

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

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

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

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

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

\(=-x+1\)

b) Để A=-6 thì -x+1=-6

\(\Leftrightarrow-x=-6-1=-7\)

hay x=7(thỏa ĐK)

Vậy: Để A=-6 thì x=7

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

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

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

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

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

b) Ta có: \(x=4+2\sqrt{3}\)

\(\Leftrightarrow x=3+2\cdot\sqrt{3}\cdot1+1\)

hay \(x=\left(\sqrt{3}+1\right)^2\)

Thay \(x=\left(\sqrt{3}+1\right)^2\) vào biểu thức \(A=\dfrac{x-1}{\sqrt{x}}\), ta được:

\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}\)

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

\(\Leftrightarrow A=\dfrac{\sqrt{3}+3}{2}\)

Vậy: Khi \(x=4+2\sqrt{3}\) thì \(A=\dfrac{\sqrt{3}+3}{2}\)

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

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

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

b: Để A<=3/căn x thì \(\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}< =\dfrac{3}{\sqrt{x}}\)

=>\(\dfrac{x-2\sqrt{x}-1-3x+6\sqrt{x}-3}{\left(\sqrt{x}-1\right)^2}< =0\)

=>\(-2x+4\sqrt{x}-4< =0\)

=>\(x-2\sqrt{x}+2>=0\)(luôn đúng)

a: 


Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

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

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

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

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

AH
Akai Haruma
Giáo viên
25 tháng 3 2021

Lời giải:

a) 

\(A=\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right].\frac{2-\sqrt{x}}{\sqrt{x}}\)

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

b) 

$A< -1\Leftrightarrow \frac{-4}{\sqrt{x}+2}+1< 0$

$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+2}< 0$

$\Leftrightarrow \sqrt{x}-2< 0\Leftrightarrow 0\leq x< 4$

Kết hợp với ĐKXĐ suy ra $0< x< 4$

14 tháng 10 2021

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

14 tháng 10 2021

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

9 tháng 7 2021

`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`

`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`

`B=2/(3-sqrtx)`

`B>1/2`

`<=>2/(3-sqrtx)-1/2>0`

`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`

`<=>(sqrtx+1)/(2(3-sqrtx))>0`

Mà `sqrtx+1>=1>0`

`<=>2(3-sqrtx)>0`

`<=>3-sqrtx>0`

`<=>sqrtx<3`

`<=>x<9`

7 tháng 12 2021

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

8 tháng 12 2021

\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế