1

\(4x^4+81\)

\(=\left(2x^2\right)^2+9^2+2.2x^2.9-2.2x^2.9\)

\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)

\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)

2

\(4a^4+b^4\)

\(=\left(2x^2\right)^2+\left(b^2\right)^2+2.2a^2.b^2-2.2a^2.b^2\)

\(=\left(2a^2+b^2\right)^2-\left(2ab\right)^2\)

\(=\left(2a^2+b^2-2ab\right)\left(2a^2+b^2+2ab\right)\)

3

\(x^7+x^5+1\)

\(=\left(x^7-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left[\left(x^3\right)^2-1^2\right]+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

\(=x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

\(=\left(x+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+x^2\left(x-1\right)+1\right]\)

4

\(x^5+x^4+1\)

\(=\left(x^5+x^4+x^3\right)+x^2+x+1-x^3-x^2-x\)

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

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

\\(C=\frac{9x^2-6x+4}{9x^2-6x+2}-1=\frac{\left(3x-2\right)^2}{9x^2-6x+2}-1\ge1\)

Do đó min C =1 khi x=2/3

 Áp dụng định lí Py-ta-go chotam giác BHC vuông tại H ta có :

\(HC^2=BC^2-HB^2\)

\(\Leftrightarrow HC^2=15^2-12^2\)

\(\Leftrightarrow HC^2=81\)

\(\Leftrightarrow HC=9\left(cm\right)\)

Ta có : \(\frac{BC}{HC}=\frac{15}{9}=\frac{5}{3}\left(1\right)\)

\(\frac{DC}{BC}=\frac{DH+HC}{BC}=\frac{16+9}{15}=\frac{25}{15}=\frac{5}{3}\left(2\right)\)

Từ (1) và (2)  \(\Rightarrow\frac{BC}{HC}=\frac{DC}{BC}\)

Xét tam giác BHC và tam giác DBC có :

\(\frac{BC}{HC}=\frac{DC}{BC}\)

Chung \(\widehat{BCD}\)

\(\Rightarrow\) tam giác BHC đồng dạng với tam giác DBC ( c-g-c )

\(\Rightarrow\widehat{DBC}=\widehat{BHC}=90^o\)

\(\Rightarrow DB\perp BC\left(đpcm\right)\)