a)xét AABC và AHBA có:

^BAC=^BHA=90

^ABC chung

-> AABC đồng dạng AHBC

b) Xét AABH và ACAH có:

^ABH=^CAH(Cùng phụ ^HAC)

^AHB=^CHA=90

-> AABH=ACAH(g-g)->\(\frac{AH}{CH}=\frac{BH}{AH}\Rightarrow AH^2=CH\cdot BH\)

c) Ta có: BC=BH + CH = 4+9=13 (cm)

-> MB=MC=\(\frac{13}{2}=6.5\) (cm)

-> HM=BM-BH= \(6,5-4=2,5\) (cm)

Ta có: \(AH^2=BH.CH\)

\(\Rightarrow AH=\sqrt{BH.CH}=\sqrt{4\cdot9}=6\) (cm)

\(\to S_{AMH}=\frac{AH.HM}{2}=\frac{6\cdot2,5}{2}=7,5\) (cm2)

The red dress is cheaper than the blue dress

(x-2)(y-5)=5

=> (x;y)=(3;10);(7;6)

\(x^2+6y^2-2xy-8x-2x+2025\)

\(=((x^2-2xy+y^2)-8\left(x-y)+16))+\left(5y^2-10y+5)+2004\right)\right.\)

\(=(x-y-4)^2+5(y-1)^2+2004\ge2004\)

dau "=" xay ra khi \(\begin{cases}x-y-4=0\\ y-1=0\Rightarrow y=1\end{cases}\)

=> x-y=4=>x-1=4=>x=5

=> GTNN M=2004 khi x=5, y=1

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

Do \(x^2\ge0\Rightarrow x^2+1\ge1\forall x\)

=> x+3=0

=> x=-3

x thuộc {2;4}, x<=3=> x=2

modern

stop

áp dụng t/c dãy tỉ số bằng nhau:

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\begin{cases}\frac{\left(a+b\right)}{c}=2\\ \frac{b+c}{a}=2\\ \frac{c+a}{b}=2\end{cases}\)

\(\begin{cases}a+b=2c\\ b+c=2a\\ c+a=2b\end{cases}\)

Thay vào ta có:

B=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}=\frac12+\frac12+\frac12=\frac32\)

3Fe+2O2--> Fe3O4

CaO+2HCl--->CaCl2+H2O

2Fe(OH)3----> Fe2O3+3H2O

SO2+2KOH---->K2SO3+H2O

2Mg+O2--->2MgO