Mọi người giúp em câu 2 và 3 với ạ
1: Thay x=9 vào A, ta được: \(A=\dfrac{3\cdot3}{3+2}=\dfrac{9}{5}\)
2: \(B=\dfrac{x+4}{x-4}-\dfrac{2}{\sqrt{x}-2}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}\)
\(=\dfrac{x+4-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
3: \(A-B< \dfrac{3}{2}\)
=>\(\dfrac{3\sqrt{x}-\sqrt{x}}{\sqrt{x}+2}< \dfrac{3}{2}\)
=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}< \dfrac{3}{2}\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+2}< \dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{4}< 0\)
=>\(\dfrac{4\sqrt{x}-3\sqrt{x}-6}{4\left(\sqrt{x}+2\right)}< 0\)
=>\(\sqrt{x}-6< 0\)
=>\(\sqrt{x}< 6\)
=>0<=x<36
mà x nguyên dương lớn nhất
nên x=35
