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\(A=\left(\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}\right)\)\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)=a-1\)
Do \(A=-a^2\Rightarrow a-1=-a^2\)=> \(a^2+a-1=0=>4a^2+4a+1-5=0=>\left(2a+1\right)^2=5\) Xét 2a+1=-5 và 5 là ra
a) \(A=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)=a-1\)
b) Ta có : \(a-1=-a^2\Leftrightarrow a^2+a-1=0\) \(\Leftrightarrow\orbr{\begin{cases}a=\frac{-1-\sqrt{5}}{2}\left(\text{loại}\right)\\a=\frac{-1+\sqrt{5}}{2}\left(\text{nhận}\right)\end{cases}}\)
2)a)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
c)\(a^3+b^3-a^2b-ab^2=a^2\left(a-b\right)-b^2\left(a-b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\\ \Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)
b)\(a^3+b^3\ge a^2b+ab^2\Leftrightarrow4a^3+4b^3\ge a^3+b^3+3a^b+3ab^2\\ \Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\Leftrightarrow\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)
a) \(ab^2\cdot\sqrt{\dfrac{3}{a^2b^4}}=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)
= \(\sqrt{3}\)
b) b. \(\sqrt{\dfrac{27\cdot\left(a-3\right)^2}{48}=}\dfrac{\sqrt{27}\cdot\sqrt{\left(a-3\right)^2}}{\sqrt{48}}\)
= \(\dfrac{3\cdot\sqrt{3}\cdot\left(a-3\right)}{\sqrt{3}\cdot\sqrt{16}}=\dfrac{3\cdot\left(a-3\right)}{4}\)
= 0.75*(a-3)
\(A=\left(a+b\right)^2+\frac{\left(ab+1\right)^2}{\left(a+b\right)^2}-2ab\)
\(A\ge2\left(ab+1\right)-2ab=2\)
\(\Rightarrow A_{min}=2\) khi \(\left(a;b\right)=\left(0;\pm1\right)\) và hoán vị