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ĐKXĐ: x>=0; x<>1

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

=>6-2x=0

=>x=3

14 tháng 7 2023

tại sao lại là 6-2x ạ

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

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

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

14 tháng 11 2023

1: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{x+3}>2\)

=>x+3>4

=>x>4-3=1

2: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

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

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

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

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

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

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

=>0<=x<4

3: ĐKXĐ: x>=0

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

=>\(x-4\sqrt{x}+3-5=x+2\sqrt{x}-5\)

=>\(x-4\sqrt{x}-2-x-2\sqrt{x}+5=0\)

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

=>\(-6\sqrt{x}=-3\)

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

=>x=1/4(nhận)

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

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

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

b: Để Q=3 thì \(\sqrt{x}-1=3\)

hay x=16

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

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

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

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

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

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

10 tháng 5 2021

B=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{6}{\sqrt{x}-1}-\dfrac{\sqrt{x}+15}{x+2\sqrt{x}-3}\)                                                         Bạn ơi giúp mk câu này vs ạ !

 

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

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

\(=\left(\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)\(=\dfrac{3\sqrt{x}-6+\sqrt{x}+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)

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

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

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

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

\(=\dfrac{3\sqrt{x}-6-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}\)

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

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

\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}=2\)

\(\Leftrightarrow x=1\)(thỏa ĐK)

27 tháng 7 2021

a) B= \(\left(\dfrac{3\left(\sqrt{x}-2\right)-1\left(\sqrt{x}+2\right)}{x-4}\right):\left(\dfrac{\sqrt{x}-6+1\left(\sqrt{x}-2\right)}{x-2\sqrt{x}}\right)\)

   \(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{2\sqrt{x}-8}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

b) Để B=\(\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\sqrt{x}\Rightarrow x=1\)

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

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

\(P=\left(\dfrac{4\sqrt{x}-6+\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right).\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)\)

\(P=\dfrac{\left(5\sqrt{x}-7\right)\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}\)

\(P=\dfrac{5\sqrt{x}-7}{2\sqrt{x}+1}\)