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

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

Hint: \(M=3-Σ\dfrac{a^2+a+1}{a^2+2a+1}\)

dự đoán dấu đẳng thức xảy ra tại \(a=b=c=\sqrt{3}-1\) nên \(MinM=\dfrac{3\sqrt{3}}{4}\)ta đi chứng minh \(M\le\dfrac{3\sqrt{3}}{4}\)

Đặt \(a+1=x\); \(b+1=y\); \(c+1=z\)

\(\Rightarrow x+y+z=a+b+c+3=a+b+c+ab+bc+ac+abc+1\)

\(=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)\(=xyz\)

\(\Rightarrow x+y+z=xyz\) đặt \(x=tan\dfrac{A}{2};y=tan\dfrac{B}{2};z=tan\dfrac{C}{2}\)

\(\Rightarrow tan\dfrac{A}{2}+tan\dfrac{B}{2}+tan\dfrac{C}{2}=tan\dfrac{A}{2}.tan\dfrac{B}{2}.tan\dfrac{C}{2}\)(đúng , tự cm hoặc google)

do \(x=tan\dfrac{A}{2}\Rightarrow sinA=\dfrac{2x}{x^2+1}\Rightarrow\dfrac{sinA}{2}=\dfrac{x}{x^2+1}\)

\(\Rightarrow M=\Sigma\dfrac{sinA}{2}\)ta phai cm \(M=\Sigma\dfrac{sinA}{2}\le\dfrac{3\sqrt{3}}{4}\Leftrightarrow M\le\dfrac{3\sqrt{3}}{2}\)

\(\Leftrightarrow sinA+sinB+sinC\le\dfrac{3\sqrt{3}}{2}\)theo BDT Cauchy-Swarch

\(\Rightarrow sinA+sinB+sinC\le\sqrt{3\left(sinA^2+sin^2B+sin^2C\right)}\)

mặt khác \(sin^2A+sin^2B+sin^2C\le\dfrac{9}{4}\)(trong sach toan 10 co BDT nay hoac google)

\(\Rightarrow sinA+sinB+sinC\le\sqrt{\dfrac{3.9}{4}}=\dfrac{3\sqrt{3}}{2}\)

\(\Rightarrow dpcm\)