\(4x^4-4x^3+4=4y^2\)

Ta có:

\(\left(2x^2-x-1\right)^2< 4x^4-4x^3+4=4y^2< \left(2x^4-x+3\right)^2\)

\(\Leftrightarrow\left(4x^4-4x^3+4\right)=\left(\left(2x^2-x\right)^2;\left(2x^2-x+1\right)^2;\left(2x^2-x+2\right)^2\right)\)

Làm nốt

\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)

\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)

\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)

\(\Leftrightarrow2\sqrt{x-8}+16=x\)

\(\Leftrightarrow x=24\)

GE≤GM+ME=12CD+12AB=AB+CD2GE≤GM+ME=12CD+12AB=AB+CD2

Dấu "=" xảy ra ⇔⇔ Ba điểm M, G, E thẳng hàng.

⇔⇔ GE // AB và GE // CD ⇔⇔ AB // CD

⇔⇔ Tứ giác ABCD là hình thang.

\(Q=\Sigma\frac{x^2}{xy^2z}+\frac{x^5}{y}+\frac{y^5}{z}+\frac{z^5}{x}\ge\frac{\left(x+y+z\right)^2}{xyz\left(x+y+z\right)}+4\sqrt[4]{\frac{x^5y^5z^5}{xyz}.\frac{1}{16}}-\frac{1}{16}\)

\(=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+2xyz-\frac{1}{16}=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+32xyz+32xyz-62xyz-\frac{1}{16}\)

\(\ge5\sqrt[5]{\frac{1}{\left(xyz\right)^2}.32^2\left(xyz\right)^2}-\frac{62}{27}\left(x+y+z\right)^3-\frac{1}{16}=20-\frac{31}{4}-\frac{1}{16}=\frac{195}{16}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)

Câu hỏi của Nguyễn Trung Hiếu - Toán lớp 9 - Học toán với OnlineMath