Question

[Đinh Quỳnh Anh]

mn giúp mình với :V

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 6:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

b: \(C=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)

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

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

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

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

c: C=4

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

=>\(\sqrt{x}=2+\dfrac{3}{4}=\dfrac{11}{4}\)

=>\(x=\left(\dfrac{11}{4}\right)^2=\dfrac{121}{16}\left(nhận\right)\)

Câu 5:

a: ĐKXĐ: x>=0 và x<>1

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

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

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

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