1: Thay x=49 vào A, ta được:
\(A=\dfrac{49+7}{7-3}=\dfrac{56}{4}=14\)
2: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{6\sqrt{x}}{9-x}-\dfrac{3}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-6\sqrt{x}-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
3: \(M=A\cdot B=\dfrac{x+7}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}\)
\(=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)
\(=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{16}{\sqrt{x}+3}}-6=2\cdot4-6=2\)
Dấu '=' xảy ra khi \(\sqrt{x}+3=4\)
=>x=1(nhận)
Giải giúp tớ với ạ, tớ cần gấp lắm
