hpt

Question

cho x,y,z dương thỏa mãn

$\begin{cases}x^{2}+xy+\frac{y^{2}}{3}=25 \\ \frac{y^{2}}{3}+z^{2}=9 \\ z^{2}+xz+x^{2}=16\end{cases}$ tính D=xy+2yz+3xz

score 5 · Answer 1

Từ điểm

$I$ cố định trong mặt phẳng, ta dựng 3 đoạn

$IA,IB,IC$ sao cho:

$\angle BIC=90^o,\angle AIB=150^o,\angle AIC=120^o$

$IA=x\sqrt3,IB=y,IC=z\sqrt3$
Kí hiệu

$S(XYZ)$ là diện tích

$\Delta XYZ$
Ta có:

$S(ABC)=S(IAB)+S(IBC)+S(ICA)$

$=\frac{1}{2}x\sqrt3.y.\sin 150^o+\frac{1}{2}y.z\sqrt3.\sin90^o+\frac{1}{2}z\sqrt3.x\sqrt3.\sin 120^o$

$=\frac{\sqrt3}{4}(xy+2yz+3zx)$
Suy ra:

$xy+2yz+3zx=\frac{4S(ABC)}{\sqrt3}$
Áp dụng định lý hàm số cosin ta có:

$AB^2=IA^2+IB^2-2IA.IB.\cos150^o=3x^2+3xy+y^2=75$

$AC^2=IA^2+IC^2-2IA.IC.\cos120^o=3x^2+3xz+3z^2=48$
Theo định lý Pytago ta có:

$BC^2=IB^2+IC^2=y^2+3z^2=27$
Suy ra:

$AB^2=AC^2+BC^2\Rightarrow \Delta ABC$ vuông tại

$C$ .

$\Rightarrow xy+2yz+3zx=\frac{4S(ABC)}{\sqrt3}=\frac{2AC.BC}{\sqrt3}=\frac{2.4\sqrt3.3\sqrt3}{\sqrt3}=24\sqrt3$

score 0 · Answer 2

Từ điểm

I cố định trong mặt phẳng, ta dựng 3 đoạn

IA,IB,IC sao cho:

∠BIC=90o,∠AIB=150o,∠AIC=120o

IA=x3√,IB=y,IC=z3√
Kí hiệu

S(XYZ) là diện tích

ΔXYZ
Ta có:

S(ABC)=S(IAB)+S(IBC)+S(ICA)

=12x3√.y.sin150o+12y.z3√.sin90o+12z3√.x3√.sin120o

=3√4(xy+2yz+3zx)
Suy ra:

xy+2yz+3zx=4S(ABC)3√
Áp dụng định lý hàm số cosin ta có:

AB2=IA2+IB2−2IA.IB.cos150o=3x2+3xy+y2=75

AC2=IA2+IC2−2IA.IC.cos120o=3x2+3xz+3z2=48
Theo định lý Pytago ta có:

BC2=IB2+IC2=y2+3z2=27
Suy ra:

AB2=AC2+BC2⇒ΔABC vuông tại

C.

⇒xy+2yz+3zx=4S(ABC)3√=2AC.BC3√=2.43√.33√3√=24

2 Đáp án