Tìm gía trị của tham số a để phương trình có nghiệm nguyên

\(\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}}=\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}=\left(\sqrt{5}+2\right).\sqrt{17-4.\sqrt{5}-8}=\left(\sqrt{5}+2\right).\sqrt{9-4\sqrt{5}}=\left(\sqrt{5}+2\right).\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}=\left(\sqrt{5}+2\right).\sqrt{\left(\sqrt{5}-2\right)^2}=\left(\sqrt{5}+2\right).\left(\sqrt{5}-2\right)=\left(\sqrt{5}\right)^2-4=5-4=1\)

Cảm ơn mọi người rất nhiều

Thế bạn là con trai à

ĐK \(x\ge-4\)

\(BPT\Leftrightarrow\hept{\begin{cases}2x-3\ge0\\x\ge-4\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{3}{2}\\x\ge-4\end{cases}}\)

\(\Rightarrow x\ge\frac{3}{2}\)

ĐK: \(x+4\ge0\) <=> \(x\ge-4\)

Bpt <=> \(\orbr{\begin{cases}x+4=0\\2x-3=0\end{cases}}\) hoặc \(2x-3>0\) <=> \(\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)hoặc \(x>\frac{3}{2}\)

<=> \(\orbr{\begin{cases}x=-4\\x\ge\frac{3}{2}\end{cases}}\)Thỏa mãn đk.

Vậy 

\(\orbr{\begin{cases}x=-4\\x\ge\frac{3}{2}\end{cases}}\)

\(\hept{\begin{cases}\left(x+y\right)^2-2xy=3\\\left(x+y\right)\left(x^2-xy+y^2\right)=27\end{cases}}\)  

Đặt S = x + y ; P = xy 

\(\hept{\begin{cases}S^2-2P=3\\S\left(S^2-2P-P\right)=27\end{cases}}\) 

   \(\hept{\begin{cases}S^2-2P=3\\S\left(3-P\right)=27\end{cases}}\)    

\(\hept{\begin{cases}S^2-2P=3\\3-P=\frac{27}{S}\end{cases}}\)   

\(\hept{\begin{cases}S^2-2\left(\frac{3S-27}{S}\right)=3\\P=\frac{3S-27}{S}\end{cases}}\)    

\(\hept{\begin{cases}S^3-6S+54=3\\P=\frac{3S-27}{S}\end{cases}}\)  

\(\hept{\begin{cases}S^3-6S+51=0\\P=\frac{3S-27}{S}\end{cases}}\)     

Tới đây giải như bình thường nha 

Ta có: \(\frac{-3}{4\sqrt{3}-7}-\frac{3}{4\sqrt{3}+7}+\sqrt{4-2\sqrt{3}}\)

\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-7^2}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

\(=\frac{-12\sqrt{3}-21-12\sqrt{3}+21}{48-49}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\frac{-24\sqrt{3}}{-1}+\left|\sqrt{3}-1\right|\)

\(=24\sqrt{3}+\sqrt{3}-1\)(vì \(\sqrt{3}>1\))

\(=25\sqrt{3}-1\)

\(=\frac{-3\left(4\sqrt{3}+7\right)-3\left(4\sqrt{3}-7\right)}{\left(4\sqrt{3}\right)^2-49}+\sqrt{4-2\sqrt{3}}\)

Tiếp tục nhé

\(A=\frac{4}{\left(x+y\right)^2}+x^2+y^2\)

\(=\frac{4}{\left(x+y\right)^2}+\left(x+y\right)^2-2xy\)

\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x+y\right)^2-2\)

\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x^2+y^2\right)+\frac{3}{4}2xy-2\)

\(=\frac{4}{\left(x+y\right)^2}+\frac{\left(x+y\right)^2}{4}+\frac{3}{4}\left(x^2+y^2\right)+\frac{3}{2}-2\)

Áp dụng bất đẳng thức Cauchy:

\(A\ge2\sqrt{\frac{4}{\left(x+y\right)^2}\frac{\left(x+y\right)^2}{4}}+\frac{3}{4}2\sqrt{x^2y^2}+\frac{3}{2}-2\)

\(A\ge2+\frac{3}{2}+\frac{3}{2}-2\)

\(A\ge3\)

\(=\left(\sqrt{3}-3\right)\cdot\sqrt{12+6\sqrt{3}}\)

\(=\left(\sqrt{3}-3\right)\cdot\sqrt{\left(\sqrt{3}+3\right)^2}\)

\(=\left(\sqrt{3}-3\right)\left(\sqrt{3}+3\right)=3-9=-6\)

\(\sqrt{2}\cdot\sqrt{6+3\sqrt{3}}=\sqrt{2\cdot\left(6+3\sqrt{3}\right)}\)

\(=\sqrt{12+6\sqrt{3}}\)