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5 tháng 8 2017

a. H=\(\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+1}\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\left(đkxđ:x\ge1\right)\)

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

b. Với x=\(\dfrac{53}{9-2\sqrt{7}}:\)

H=\(-2\sqrt{\dfrac{53}{9-2\sqrt{7}}-1}+\dfrac{53}{9-2\sqrt{7}}\)

H\(=7\)

c. \(-2\sqrt{x-1}+x=16\)

\(\sqrt{x-1}=\dfrac{x-16}{2}\)

\(4x-4=x^2-32x+256\)

\(x^2-36x+260=0\)

x=26

d. Để H>1 thì x>3

5 tháng 8 2017

bạn làm rõ ràng hơn đc không

8 tháng 8 2018

a) Rut gon H

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

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

DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)

Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

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

\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)

a: \(H=\dfrac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{x-1-x}+x\)

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

b: \(x=\dfrac{53}{9-2\sqrt{7}}=9+2\sqrt{7}\)

Khi x=9+2 căn 7 thì \(H=-2\cdot\sqrt{8+2\sqrt{7}}+9+2\sqrt{7}\)

\(=-2\left(\sqrt{7}+1\right)+9+2\sqrt{7}\)

=-2+9=7

a: ĐKXĐ: x>0; x<>4

\(P=\left(2-\sqrt{x}+2\right)\cdot\dfrac{1}{\sqrt{x}-2}=\dfrac{4-\sqrt{x}}{\sqrt{x}-2}\)

b: P=2/3

=>(4-căn x)/(căn x-2)=2/3

=>2căn x-4=12-3căn x

=>5căn x=16

=>x=256/25

c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)

a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:

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

c: Ta có: \(P< \dfrac{1}{2}\)

\(\Leftrightarrow P-\dfrac{1}{2}< 0\)

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

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

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

19 tháng 8 2017

A=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

ĐKXĐ :x\(\ne\)9,x\(\ge\)0

<=> \(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\)

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

=\(\dfrac{3\sqrt{x}-9}{x-9}\)=\(\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)

ta có : A=1/3 => \(\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}=>\sqrt{x}+3=9\)

=> x=36

vậy giá trị của x=36 khi A=1/3

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

ĐKXĐ: x\(\ne\)1 ,x\(\ge\)0

<=> \(\dfrac{\sqrt{x}+1-\sqrt{x}}{x-1}:\dfrac{1}{\sqrt{x}+1}\)

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

ta có : B<0 =>\(\dfrac{1}{\sqrt{x}-1}< 0\)

=> 1< \(\sqrt{x}-1\)=> \(\sqrt{x}\)>2=>x>4

vậy x>4 khi B<0

18 tháng 8 2017

\(ĐKXĐ:x\ne9\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9} =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{3x+9}{x-9}=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)

30 tháng 10 2018

a) điều kiện : \(x>0;x\ne9\)

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

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

\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\) \(\Leftrightarrow A=\left(\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\right)\) \(\Leftrightarrow A=\dfrac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\) b) \(A< -1\) \(\Leftrightarrow A+1< 0\Leftrightarrow\dfrac{-3\sqrt{x}}{2\sqrt{x}+3}+1< 0\Leftrightarrow\dfrac{-\sqrt{x}+3}{2\sqrt{x}+3}< 0\)

\(\Leftrightarrow-\sqrt{x}+3< 0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

vậy ...

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29 tháng 10 2018

Mysterious Person giup mk nha. Mk cảm ơn!