canhhoang30011999

câu 1b

MO ko la phân giác CMN

$\bigtriangleup MOB\sim \bigtriangleup ONC$

$\Rightarrow \frac{OM}{ON}= \frac{OB}{NC}= \frac{OC}{NC}$

$\Rightarrow \frac{OM}{OC}= \frac{ON}{NC}$

$\angle NCO= \angle MON= 60$

$\Rightarrow \bigtriangleup CON\sim \bigtriangleup OMN$

$\Rightarrow \angle MNO= \angle ONC$

=>NO là phân giác $\angle MNC$

câu 1a

$\angle MOB+\angle OBM+\angle BMO= 180$

$\angle MOB+\angle MON+\angle NOC= 180$

$\angle MBO= \angle MON= 60$

$\Rightarrow \angle BMO= \angle NOC$

$\Rightarrow \bigtriangleup BMO\sim \bigtriangleup CON$

$\Rightarrow \frac{OB}{NC}= \frac{MB}{OC}$

$\Rightarrow OB*OC= BM*NC$

$\Rightarrow BC^{2}= 4BM*NC$

cau a

$\angle ACN= \angle ABM$

$\frac{BM}{MH}= \frac{CN}{ND}$$\Rightarrow \frac{CN}{CD}= \frac{BM}{BH}$$\Rightarrow \frac{CN}{BM}= \frac{CD}{BH}$

tam giac ABH dong dang tam giac DBA

=>$\frac{AB}{DB}= \frac{BH}{AB}= \frac{BH}{DC}= \frac{AB}{AC}$

=>$\frac{CN}{BM}= \frac{CD}{BH}= \frac{AC}{AB}$

tam giacNCA dong dang tam giacMBA(cgc)

1actheo dl Pytago ta co dpcm

caub sai de

cau c $\angle BDA+\angle ABD= 90$

$\angle AID= \angle BIH$

$\angle BIH+\angle DBC= 90$

$\Rightarrow \angle AID= \angle ADI$

=>TAM GIAC AID can o A

1 gio nguoi thu 1 ,2,3$\frac{1}{8}$$\frac{1}{10}$lam duoc$\frac{1}{14}$

trong 2h30ph dau 2ng lam duoc $\left ( \frac{1}{10}+\frac{1}{14} \right )2.5= \frac{3}{7}$ cong viec

=>con lai $\frac{4}{7}$ cong viec

nguoi thu 1 lam $\frac{4}{7}\div \left ( \frac{1}{8}+\frac{1}{10} \right )= \frac{20}{21}$ gio

cau b

theo dl ta let ta co

$\frac{QM}{AH}= \frac{BQ}{AB}$

$\frac{QP}{BC}= \frac{AQ}{AB}$$\Rightarrow \frac{QM}{AH}+\frac{QP}{BC}= 1$

MA AH=BC$\Rightarrow QM+QP= BC\Rightarrow Pmnpq= 2BC$ KHONG DOI

bai 5

A$= 3\left ( n^{2}+2n+1 \right )+10$$= 3\left ( n+1 \right )^{2}+10$

do n le suy ra n+1 chan

suy ra $3\left ( n+1 \right )^{2}$ chia het cho 4

ma 10 chia 4 du 2

=>A chia 4 du 2=>A khong la scp

bai 2b

dat x=1-a,y=2+a(a>0)

bdt tuong duong voi

$\left ( a+2 \right )^{3}-\left ( 1-a \right )^{3}-6\left ( a+2 \right )^{2}-\left ( 1-a \right )^{2}+9\left ( a+2 \right )\geq 0$

$\Leftrightarrow 2a^{3}-4a+2a\geq 0$$\Leftrightarrow 2a\left ( a^{2} -2a+1\right )\geq 0$$a\geq 0$$\left ( a-1 \right )^{2}\geq 0$$\Rightarrow dpcm$

dau bang xay ra khi a=0 hoac a=1

a=o =>x=1,y=2

a=1=>x=0,y=3

 bai 1

max$y= \frac{2x+m}{x^{2}+1}= 3\Leftrightarrow \frac{2x+m}{x^{2}+1}\leq 3$$\Leftrightarrow 2x+m\leq 3(x^{2}+1)$$\Leftrightarrow 2x+m\leq 3(x^{2}+1)$

$3x^{2}-2x+3-m\geq 0$$\Leftrightarrow 1-3\left ( 3-m \right )\leq 0$$\Leftrightarrow m\leq \frac{8}{3}$

bất đẳng thức tương đương 

$\sum \frac{1}{-a^{2}-b^{2}-2}\geq \frac{-3}{4}$

sư dụng bdt cauchy-schwart ta có

$\sum \frac{1}{-a^{2}-b^{2}-2}\geq \frac{9}{-2\left ( a^{2}-b^{2}-c^{2} \right )-6}$

$\geq \frac{9}{2\left ( -3 \right )-6}= \frac{-3}{4}$