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PP
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PM
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H24
11 tháng 3 2017 lúc 20:28

A B M C

a) M điểm giữa của BC nên CM = MB

Tam giác ABM và tam giác ACM có : đáy CM = MB, chung đường cao tương ứng với đáy.

Nên S(ABM) = S(ACM)

b) S(ABC) = S(ABM) + S(ACM) mà S(ABM) = S(ACM)

Nên  \(S\left(ABM\right)=\frac{1}{2}S\left(ABC\right)\)

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PM
13 tháng 3 2017 lúc 11:28

1/2 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000% luob

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PM
20 tháng 4 2017 lúc 19:54

1/2 dung 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000% luon minh the

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TC
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LT
19 tháng 8 2017 lúc 21:12

abc-bca

=100a+10b+c-100c-10b-a

=( 100a-a )+(10b-10b)-(100c-c)

=99a-99c=99,(a-c)chia hết cho 99 

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MT
19 tháng 8 2017 lúc 21:13

abc - bca 

=  a.100 + b.10 + c - b.100 - c.10 - a

= ( a.100 -a ) + ( 10b - 10b ) - ( 100c - c )

= a.99 - c.99

= 99.(a-c) chia hết cho 99

Vậy abc - bca chia hết cho 99

Nhớ k cho mình nhé! Thank you!!!

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TM
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HL
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VH
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TH
14 tháng 2 2016 lúc 16:48

moi hok lop 6

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VN
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NM
29 tháng 3 2019 lúc 13:06

Theo đề bài

\(\overline{abc,d}-\overline{a,bcd}=583,704\)

\(\Rightarrow100x\overline{a,bcd}-\overline{a,bcd}=583,704\)

\(\Rightarrow99x\overline{a,bcd}=583,704\Rightarrow\overline{a,bcd}=583,704:99=5,896\Rightarrow\overline{abc,d}=589,6\)

Tổng hai số là

\(\overline{589,6}+5,896=595,496\)

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TN
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LG
5 tháng 3 2017 lúc 13:18

B A C D

Xét \(\Delta ABC\)\(\Delta ABD\) ,có :

AB : cạnh chung

AC =AD (gt)

\(\widehat{BAC}=\widehat{BAD}=90^0\)

=> \(\Delta ABC=\Delta ABD\) ( hai cạnh góc vuông )

=> BC =BD ( 2 cạnh tương ứng )

=> \(\Delta BCD\) cân tại B

Ta có : AC = 1/2 BC

Lại có : AD =AC => AD + AC = BC hay DC = BC = BD

=> \(\Delta BCD\) là tam giác đều

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ND
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KT
21 tháng 7 2018 lúc 20:17

\(abc+ac=2.ca\)   (1)

  \(a=0\)hoặc \(c=0\)là nghiệm của  (1)\(a,c\ne0\)thì:   \(\frac{abc+ac}{ca}=\frac{2ca}{ca}\)

                           \(\Leftrightarrow\)\(b+1=2\) \(\Leftrightarrow\)\(b=1\)

Vậy....

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HH
21 tháng 7 2018 lúc 20:21

\(abc-ac=2.ca\left(1\right)\)

* a = 0 hoặc c = 0 là nghiệm của ( 1 )

* a.c \(\ne\) thì \(\frac{abc+ac}{ca}=\frac{2ca}{ca}\)

\(\Leftrightarrow b+1=2\Leftrightarrow b=1\)

\(KL:\)

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ND
21 tháng 7 2018 lúc 20:21

co cach nao de hieu ko chi?

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