Question

Mọi giúp mik câu 13 nhé

Answer 1

a: \(A=\dfrac{3x+5\sqrt{x}-11}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+2}-1\)

\(=\dfrac{3x+5\sqrt{x}-11}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+2}-1\)

\(=\dfrac{3x+5\sqrt{x}-11-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-1\)

\(=\dfrac{3x+5\sqrt{x}-11-x+4+2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-1\)

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

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

\(=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\)

b: A=2

=>\(2\left(\sqrt{x}+2\right)=\sqrt{x}+7\)

=>\(2\sqrt{x}+4=\sqrt{x}+7\)

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

=>x=9(nhận)