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\(\widehat{B_1}\) = \(\widehat{B_2}\) = 1000 (hai góc đối đỉnh)
\(\widehat{C_2}\) = \(\widehat{B_1}\) = 1000 (hai góc đồng vị)
\(\widehat{C_3}\) + \(\widehat{C_2}\) = 1800 ⇒ \(\widehat{C_3}\) = 1800 - 1000 = 800
\(\widehat{D_1}\) = \(\widehat{A_1}\) = 600 (so le trong)
\(\widehat{DAH}\) = 900 - 600 = 300
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a. \(A_4=B_2=37^0\left(slt\right)\)
b. \(A_1=B_1\left(dongvi\right)\)
c. \(B_3=180^0-B_2=180^0-37^0=143^0\left(kebu\right)\)
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a, Vì a//b và b⊥c nên a⊥c
b, Ta có \(\widehat{D_2}=\widehat{D_4}=65^0\) (đối đỉnh)
Vì a//b nên \(\widehat{C_4}=\widehat{D_2}=65^0\) (so le trong)
\(\widehat{C_3}+\widehat{C_4}=180^0\) (kề bù)
Hay \(\widehat{C_3}=180^0-65^0=115^0\)
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B1 : S = 1 + 2 + 2^2 + 2^3 + ... + 2^2008 / 1 - 2^2009
Đặt A = 1 + 2 + 2^2 + 2^3 + ... + 2^2008
2A = 2 + 2^2 + 2^3 + 2^3 + 2^4 + ... + 2^2009
2A - A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^2009 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^2008 )
A = 2^2009 - 1
S = 2^2009 - 1 / 1 - 2^2009
S = -1
Ta có: AB//DC
=>\(\widehat{A_1}=\widehat{D_4}\)(hai góc so le trong)
=>\(\widehat{D_4}=110^0\)
Ta có: \(\widehat{D_1}=\widehat{D_4}\)(hai góc đối đỉnh)
mà \(\widehat{D_4}=110^0\)
nên \(\widehat{D_1}=110^0\)
Ta có: AB//DC
=>\(\widehat{C_3}=\widehat{B_2}\)(hai góc so le trong)
=>\(\widehat{B_2}=135^0\)
Ta có: \(\widehat{B_1}+\widehat{B_2}=180^0\)(hai góc kề bù)
=>\(\widehat{B_1}=180^0-135^0=45^0\)