`Answer:`

Bài 8:

`N=2x^4+3x^2y^2+x^4+y^2`

`=2x^4+2x^2y^2+x^2y^2+x^4+y^2`

`=(2x^4+2x^2y^2)+(x^2y^2+x^4)+y^2`

`=2x^2.(x^2+y^2)+x^2.(x^2+y^2)+y^2`

Mà đề ra `x^2+y^2=1`

\(\Rightarrow N=2x^2.1+x^2.1+y^2=2x^2+x^2+y^2=2x^2+\left(x^2+y^2\right)=2x^2+1\)

Bài 9:

Theo đề ra `(x^2+2010)(x-10)=0`

Dễ thấy `x^2+2010>=2010>0<=>x-10=0`

`P=(x^2-1)(x^2-2)(x^3-3)...(x^2-2015)`

`=(x^2-1)...(x^2-100)...(x^2-2015)`

`=(x^2-1)...(x-10)(x+1)...(x^2-2015)`

`=0`

Bài 10:

`M=2x+2y+3xy(x+y)+5x^2y^3+5x^3y^2+2`

`=2(x+y)+3xy(x+y)+5x^3y^2+5x^2y^3+2`

`=2(x+y)+3xy(x+y)+5x^2y^2(x+y)+2`

`=(x+y)(2+3xy+5x^2y^2)+2`

Thay `x+y=0` vào biểu thức `M`, ta được: \(M=0\left(2+3xy+5x^2y^2\right)+2=2\)

Bài 11:

`a)` `P=-2(x-3)^2+5`

Xét `-2(x-3)^2<=0`

`=>P<=0+5=5`

Dấu "=" xảy ra khi `x=3`

`b)` `Q=\frac{5}{(x-14)^2+21}`

Xét `(x-14)^2+21>=21`

`=>\frac{1}{(x-14)^2+21}<=\frac{1}{21}`

`=>Q<=\frac{5}{21}`

Dấu "=" xảy ra khi `x=14`

\(\frac{2^{12}.243-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3-1\right)}\)

\(=\frac{1}{3}\)

HT

a.Ta có \(Δ A S P\) vuông tại S
\(→ A P ^2 = S A ^2 + S P ^2 = 225\)

\(→ A P = 15 \)

b.Xét\(Δ K A H , Δ G P H\)có:

\(ˆ A K H = ˆ P G H ( = 90 ^o )\)

\(H A = H P\) vì H là trung điểm AP

\(ˆ K H A = ˆ P H G\)

\(→ Δ K A H ∼ Δ G P H\)(cạnh huyền-góc nhọn)

c.Từ câu a\(→ H K = H G → H\) là trung điểm KG

Ta có:

\(K H ^2 = A H ^2 − A K ^2\)

\(→ 4 K H ^2 = 4 A H ^2 − 4 A K ^2\)

\(→ ( 2 K H ) ^2 = ( 2 A H ) ^2 − 4 K A ^2\)

\(→ K G ^2 = A P ^2 − 4 K A ^2\)

HT