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\(a,A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{2^{2018}}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}+\dfrac{1}{3^{2017}}\)
\(3A-A=1-\dfrac{1}{3^{2018}}\)
\(A=\dfrac{\left(1-\dfrac{1}{3^{2018}}\right)}{2}\)
\(b,B=1+5+5^2+5^3+...+5^{100}\)
\(5B=5+5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=1-5^{101}\)
\(B=\dfrac{\left(1-5^{101}\right)}{4}\)
a) Vì M, B thuộc 2 tia đối nhau CB và CM
=> C nằm giữa B và M
=> BM = BC + CM =8 (cm)
b) Vì C nằm giữa B, M
=> Tia AC nằm giữa tia AB và tia AM
=> góc CAM = góc BAM - góc BAC = 20 độ
c) Ta có :
Góc xAy = góc xAC + góc CAy = 1/2 góc BAC + 1/2 góc CAM
= 1/2 (góc BAC + góc CAM) = 1/2 góc BAM 1/2 x 80 độ = 40 độ
a,
\(\left(20+9\dfrac{1}{4}\right):2\dfrac{1}{4}=\left(20+\dfrac{37}{4}\right):\dfrac{9}{4}\\ =\dfrac{117}{4}\cdot\dfrac{4}{9}\\ =\dfrac{117}{9}=13\)
b,
\(\left(6-2\dfrac{4}{5}\right)\cdot3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\\ =\left(6-\dfrac{14}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\\ =\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\\ =10-\dfrac{32}{5}\\ =\dfrac{18}{5}\)
c,
\(\dfrac{32}{15}:\left(-1\dfrac{1}{5}+1\dfrac{1}{3}\right)\\ =\dfrac{32}{5}:\left(-\dfrac{6}{5}+\dfrac{4}{3}\right)\\ =\dfrac{32}{5}:\dfrac{2}{15}\\ =\dfrac{32}{5}\cdot\dfrac{15}{2}\\ =48\)
a, ( 20 + \(9\dfrac{1}{4}\) ) : \(2\dfrac{1}{4}\)
= ( 20 + \(\dfrac{37}{4}\) ) : \(\dfrac{9}{4}\)
= ( \(\dfrac{80}{4}\) + \(\dfrac{37}{4}\) ) . \(\dfrac{4}{9}\)
= \(\dfrac{117}{4}\) . \(\dfrac{4}{9}\)
= \(\dfrac{117}{9}\) = 13
b, ( 6 - \(2\dfrac{4}{5}\) ) . \(3\dfrac{1}{8}\) - \(1\dfrac{3}{5}\) : \(\dfrac{1}{4}\)
= ( 6 - \(\dfrac{14}{5}\) ) . \(\dfrac{25}{8}\) - \(\dfrac{8}{5}\) . 4
= ( \(\dfrac{30}{5}\) - \(\dfrac{14}{5}\) ) . \(\dfrac{25}{8}\) - \(\dfrac{8}{5}\) . 4
= \(\dfrac{16}{5}\) . \(\dfrac{25}{8}\) - \(\dfrac{8}{5}\). 4
= 10 - \(\dfrac{32}{5}\)
= \(\dfrac{50}{5}\) - \(\dfrac{32}{5}\)
= \(\dfrac{18}{5}\)
c, \(\dfrac{32}{15}\) : ( -\(1\dfrac{1}{5}\) + \(1\dfrac{1}{3}\) )
= \(\dfrac{32}{15}\) : ( \(\dfrac{-6}{5}\) + \(\dfrac{4}{3}\) )
= \(\dfrac{32}{15}\) : ( \(\dfrac{-18}{15}\) + \(\dfrac{20}{15}\) )
= \(\dfrac{32}{15}\) : \(\dfrac{2}{15}\)
= \(\dfrac{32}{15}\) . \(\dfrac{15}{2}\)
= 16
\(\frac{1}{-4}-\frac{4}{-3}+\frac{1}{-3}\left(\frac{50}{100}-\frac{5}{2}\right)=-\frac{1}{4}+\frac{4}{3}-\frac{1}{3}\left(\frac{1}{2}-\frac{5}{2}\right)=-\frac{1}{4}+\frac{4}{3}-\frac{1}{3}\left(-\frac{4}{2}\right)\)
\(=-\frac{1}{4}+\frac{4}{3}+\frac{2}{3}=-\frac{1}{4}+\frac{6}{3}=-\frac{1}{4}+2=-\frac{1}{4}+\frac{8}{4}=\frac{7}{4}\)
\(-\frac{14}{10}.\frac{15}{-49}-\frac{6}{3}:\frac{13}{5}=\frac{7.2}{2.5}.\frac{3.5}{7.7}-2.\frac{5}{13}=\frac{3}{7}-\frac{10}{13}=\frac{39}{91}-\frac{70}{91}=-\frac{31}{91}\)
3)
A B I K
a) Vì (A; R=3 cm) cắt AB tại K
=> K nằm trên đường tròn (A; 3 cm)
=> AK=3 cm
Vì (B; 2 cm) cắt AB=I
=> I nằm trên đường tròn (B; 2 cm)
=> BI=2cm
b) Có: AI=AB--BI=4-2=2cm
IK=AK-AI=3-2=1 cm
=>AI>IK
c) KB=BI-IK=2-1=1 cm
=> KB=IK
I, K, B thẳng hàng
=> K là trung điểm IB
\(\frac{1}{-4}-\frac{4}{-3}+\frac{1}{-3}.\left(50\%-1\frac{3}{2}\right)\)
=\(\frac{-1}{4}-\frac{-4}{3}+\frac{-1}{3}.\left(50\%-\frac{5}{2}\right)\)
=\(\frac{-1}{4}-\frac{-4}{3}+\frac{-1}{3}.\left(-2\right)\)
=\(\frac{-1}{4}-\frac{-4}{3}+\frac{2}{3}\)
=\(\frac{-1}{4}+\frac{4}{3}+\frac{2}{3}\)
=\(\frac{-1}{4}+\left(\frac{4}{3}+\frac{2}{3}\right)\)
=\(\frac{-1}{4}+2\)
=\(\frac{7}{4}=1,75\)
\(-1,4.\frac{15}{-49}-\left(\frac{2}{3}+\frac{4}{3}\right):2\frac{3}{5}\)
=\(-1,4.\frac{15}{-49}-2:2\frac{3}{5}\)
=\(\frac{-1}{4}.\frac{15}{-49}-\frac{2}{1}:\frac{13}{5}\)
=\(\frac{15}{196}-\frac{10}{13}\)
=\(\frac{-1765}{2548}\)
MIK KO VẼ ĐC TRÊN NÀY, SORRY.
a) KA= bán kính đường tròn tâm A = 3cm
IB= bán kính đường tròn tâm B= 2cm
b) AI= AB- bán kính đường tròn tâm B
= 4cm-2cm
=2cm
IK= AB-AI-KB
= 4cm- 2cm- (AB-AK)
= 4cm-2cm-(4cm-3cm)
= 4cm-2cm-1cm
= 1cm
=> AI>IK
c) KB=AB- AK
= 4cm-3cm
=1cm
Vì K nằm giữa I và B và IK=KB=1cm
=> K là trung điểm của đoạn thẳng IB
k cho mik nha
HELLO
cái này là hỗn số bạn nhé 2+1/2 ban