Question

Giúp mình vs mình cần gấp lắm ạ

Answer 1

Bài I:

1: Thay x=4 vào A, ta được:

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

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

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

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

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

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

3: P=A*B

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

P<=4

=>P-4<=0

=>\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-1}< =0\)

=>\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}< =0\)

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

=>\(\sqrt{x}< 1\)

=>0<=x<1

Kết hợp ĐKXĐ, ta được: 0<=x<1