Cho tg ABC cân tại A, có góc A < 90*, kẻ BH vuông góc với AC, CK vuông góc với AC. gọi O là giao điểm của BH và CK.
a, cm tg ABH= tg ACH.
b,tg OBC cân
c, tg OBK=tgOCK
đặt a = AB = AC
Áp dụng định lý pytogo trong tam giác vuông ta có
\(a^2+a^2=BC^2\Rightarrow2a^2=12^2=144\Rightarrow a^2=72\Leftrightarrow a=\sqrt{72}=6\sqrt{2}\)
vậy, AB = AC = \(6\sqrt{2}\)
\(\dfrac{-22}{45}=-\dfrac{2266}{45.103}>-\dfrac{2295}{45.103}=-\dfrac{51}{103}\)
Ta có: \(\dfrac{-22}{45}=\dfrac{-22\cdot103}{45\cdot103}=\dfrac{-2266}{4635}\)
\(\dfrac{-51}{103}=\dfrac{-51\cdot45}{103\cdot45}=\dfrac{-2295}{4635}\)
mà -2266>-2295
nên \(\dfrac{-22}{45}>\dfrac{-51}{103}\)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
Ta có :
+) \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}\)
+) \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}\)
\(\Leftrightarrow S< \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{40}+\dfrac{1}{40}\)
\(\Leftrightarrow S< \dfrac{1}{2}\)
Vậy,,,
Ta có: \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{2}{40}=\dfrac{1}{20}\)
Do đó: \(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{4}+\dfrac{1}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{1}{2}\)
hay \(S< \dfrac{1}{2}\)(đpcm)
Ta có: \(32^{27}=\left(2^5\right)^{27}=2^{135}\)
\(16^{39}=\left(2^4\right)^{39}=2^{156}\)
mà \(2^{135}< 2^{156}\)
nên \(32^{27}< 16^{39}\)
mà \(16^{39}< 18^{39}\)
nên \(32^{27}< 18^{39}\)
\(\Leftrightarrow-32^{27}>-18^{39}\)
\(\Leftrightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)
16 = 24
(\(\dfrac{1}{16}\))200 = \(\dfrac{1}{2^{4.200}}\) = \(\dfrac{1}{2^{800}}\)= (\(\dfrac{1}{2}\))800
So sánh với (\(\dfrac{1}{2}\))1000
Hai phân số cùng tử số, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn
Suy ra: (\(\dfrac{1}{16}\))200 > (\(\dfrac{1}{2}\))1000
Ta có: \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}\)
mà \(\left(\dfrac{1}{2}\right)^{800}>\left(\dfrac{1}{2}\right)^{1000}\)
nên \(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)
Nếu \(a>b\Rightarrow an>bn\Rightarrow ab+an>ab+bn\)
\(\Leftrightarrow a\left(b+n\right)>b\left(a+n\right)\)
\(\Leftrightarrow\dfrac{a+n}{b+n}< \dfrac{a}{b}\)
Nếu \(a< b\Rightarrow an< bn\Rightarrow ab+an< ab+bn\)
\(\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)
\(\Leftrightarrow\dfrac{a+n}{b+n}>\dfrac{a}{b}\)
Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
Vì \(10^{11}-1< 10^{12}-1\)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)