Chứng minh
0+a=a+o=a
a+b=b+a
(a+b)+c=a+(b+c)
là đúng
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NL
Nguyễn Lê Phước Thịnh
CTVHS
14 tháng 12 2023
a: Xét tứ giác ABOC có
\(\widehat{ABO}+\widehat{ACO}=90^0+90^0=180^0\)
=>ABOC là tứ giác nội tiếp
=>A,B,O,C cùng thuộc một đường tròn
Xét (O) có
AB,AC là các tiếp tuyến
Do đó: AB=AC
=>A nằm trên đường trung trực của BC(1)
Ta có: OB=OC
=>O nằm trên đường trung trực của BC(2)
Từ (1) và (2) suy ra OA là đường trung trực của BC
=>OA\(\perp\)BC tại H
b: Xét (O) có
ΔEDB nội tiếp
BD là đường kính
Do đó: ΔBED vuông tại E
=>BE\(\perp\)ED tại E
=>BE\(\perp\)AD tại E
Xét ΔDBA vuông tại B có BE là đường cao
nên \(AE\cdot AD=AB^2\left(3\right)\)
XétΔABO vuông tại B có BH là đường cao
nên \(AH\cdot AO=AB^2\left(4\right)\)
Từ (3) và (4) suy ra \(AE\cdot AD=AH\cdot AO\)
15 tháng 6 2022
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
17 tháng 4 2022
\(\overrightarrow{BD'}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DD'}=-\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}\)
\(=-\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\)
LD
7 tháng 10 2016
Áp dụng tính chất dãy tỷ số bằng nhau, ta có:
a/b=b/c=c/a=a+b+c/b+c+a=1
⇒a/b=b/c=c/a=1
=> a = b = c = 2013
Vậy b = 2013; c = 2013
a/b=b/c=c/a=a+b+c/b+c+a=1
⇒a/b=b/c=c/a=1
=> a = b = c = 2013
Vậy b = 2013; c = 2013
NL
Nguyễn Lê Phước Thịnh
CTVHS
13 tháng 2
a: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\)
\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\)
Do đó: \(\frac{a}{a-b}=\frac{c}{c-d}\)
c: \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\)
Do đó: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
e: \(\frac{a\cdot b}{c\cdot d}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)