ĐKXĐ: -2x+1>=0

=>-2x>=-1

=>x<=1/2

- Ta có: \(A=\frac{\sqrt{x+1}}{\sqrt{x-1}}\)

- Thay  \(x=\frac{16}{9}\)vào đa thức \(A,\)ta có:

             \(A=\frac{\sqrt{\frac{16}{9}+1}}{\sqrt{\frac{16}{9}-1}}\)

      \(\Leftrightarrow A=\frac{\sqrt{\frac{25}{9}}}{\sqrt{\frac{7}{9}}}\)

      \(\Leftrightarrow A=\frac{5\sqrt{7}}{7}\)

Vậy \(A=\frac{5\sqrt{7}}{7}\)

Thay x = 16/9 vào biểu thức, ta có: 

\(\frac{\sqrt{\frac{16}{9}+1}}{\sqrt{\frac{16}{9}-1}}=\frac{\sqrt{\frac{25}{9}}}{\sqrt{\frac{7}{9}}}=\frac{\frac{5}{3}}{\frac{\sqrt{7}}{3}}=\frac{5\sqrt{7}}{5}\)

\(\left\{{}\begin{matrix}x+y=2\left(m-1\right)\left(1\right)\\2x-y=m+8\left(2\right)\end{matrix}\right.\) 

Cộng từng vế của (1) và (2) ta được: 

\(3x=3m+6=3\left(m+2\right)\) \(\Leftrightarrow x=m+2\) Thay vào (2) ta được:

\(\Rightarrow2\left(m+2\right)-y=m+8\) \(\Leftrightarrow y=2m+4-m-8=m-4\)

\(\Rightarrow x^2+y^2=\left(m+2\right)^2+\left(m-4\right)^2=m^2+4m+4+m^2-8m+16=2m^2-4m+20=2m^2-4m+2+18=2\left(m^2-2m+1\right)+18=2\left(m-1\right)^2+18\ge18\)

GTNN của \(x^2+y^2=18\Leftrightarrow m=1\)

1.

\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)

Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)

Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)

Từ đó ta được đpcm

2.

\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)

\(c,=a^2b^2\left(b-a\right)+b^2c^2\left(c-a+a-b\right)-c^2a^2\left(c-a\right)\\ =-a^2b^2\left(a-b\right)+b^2c^2\left(a-b\right)+b^2c^2\left(c-a\right)-c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(b^2c^2-a^2b^2\right)+\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(c-a\right)\left(c+a\right)+c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(c-a\right)\left[b^2\left(c+a\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b^2c+ab^2-bc^2-ac^2\right)\\ =\left(a-b\right)\left(c-a\right)\left[bc\left(b-c\right)+a\left(b-c\right)\left(b+c\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(bc+ab+ac\right)\)

1: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)

=>(x+y)^2>=4xy

=>(x-y)^2>=0(luôn đúng)

2: \(\Leftrightarrow a^3+b^3-a^2b-ab^2>=0\)

=>a^2(a-b)-b^2(a-b)>=0

=>(a-b)^2(a+b)>=0(luôn đúng)