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

b: Để P=-1 thì \(\sqrt{x}-1=-\sqrt{x}\)

=>x=1/4(nhận)

Bài 1:

a: ĐKXĐ: 2x+3>=0 và x-3>0

=>x>3

b: ĐKXĐ:(2x+3)/(x-3)>=0

=>x>3 hoặc x<-3/2

c: ĐKXĐ: x+2<0

hay x<-2

d: ĐKXĐ: -x>=0 và x+3<>0

=>x<=0 và x<>-3

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

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

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

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

\(\Leftrightarrow2x+4\sqrt{x}+2-\sqrt{x}=0\)

\(\Leftrightarrow2x+3\sqrt{x}+2=0\)(1)

Đặt \(\sqrt{x}=a\)(a>=0)

(1) trở thành \(2a^2+3a+2=0\)

\(\Delta=3^2-4\cdot2\cdot2=9-16=-7< 0\)

Do đó: (1) vô nghiệm

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

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

b: Để P<0 thì x-1<0

hay 0<x<1

19 tháng 10 2018

1.

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b,

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

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

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

\(=(\sqrt{x}\times2\sqrt{x})\times\dfrac{1}{2\sqrt{x}}\)

\(=\sqrt{x}\)

c,

\(M>3\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

25 tháng 10 2022

Bài 2: 

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

b: Để A=căn A thì A=1 hoặc A=0

=>A=1

=>1-a=1

=>a=0

c: Thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\) vào A, ta được:

\(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)

16 tháng 8 2018

Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)

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

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

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

16 tháng 8 2018

Bài 3:

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

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

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

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

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

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

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

27 tháng 11 2018

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

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

\(Q=x+1\)

Không thể tìm được GTLN hay GTNN của Q.

b)

   \(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)

Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)

Vậy x=1, x=9 là các giá trị cần tìm

23 tháng 9 2018

ĐK:x>0,x≠0,x≠1

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{x-1}\right)=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{x-\sqrt{x}}\)b) Khi x=\(3+2\sqrt{2}\) thì \(P=\dfrac{\sqrt{3+2\sqrt{2}}-3}{3+2\sqrt{2}-\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}+1}-3}{3+2\sqrt{2}-\sqrt{2+2\sqrt{2}+1}}=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-3}{3+2\sqrt{2}-\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}+1-3}{3+2\sqrt{2}-\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{2+\sqrt{2}}=\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\)

c) Ta có \(P< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{x-\sqrt{x}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\x-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\x-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1< x< 9\)

Vậy 1<x<9 thì P<0

24 tháng 9 2018

tại sao lại suy ra được 1<x<9 vậy

bạn giải thích giùm mình với

11 tháng 2 2019

1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1

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

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

b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)

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

\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))

\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)

\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)