C là trung điểm của OA

=>\(OC=\dfrac{OA}{2}=4,5\left(cm\right)\)

Trên tia Ox, ta có: OC<OB

nên C nằm giữa O và B

=>OC+CB=OB

=>CB+4,5=12

=>CB=12-4,5=7,5(cm)

a) \(A=1+3+3^2+...+3^{100}\)

\(3A=3+3^2+3^3+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\frac{3^{101}-1}{2}\)

b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)

\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)

\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)

\(3B=2^{101}+1\)

\(B=\frac{2^{101}+1}{3}\)

a) $25.12$

$=25.4.3$

$=100.3=300$

b) $34.99$

$=34.(100-1)$

$=34.100-34.1$

$=3400-34=3366$

c) $47.101$

$=47.(100+1)$

$=47.100+47.1$

$=4700+47=4747$

d) $15.302$

$=15.(300+2)$

$=15.300+15.2$

$=4500+30=4530$

e) $125.32$

$=125.8.4$

$=1000.4=4000$

g) $123.1001$

$=123.(1000+1)$

$=123.1000+123.1$

$=123000+123=123123$

Bạn tự vẽ hình nhé

Ta có E là trung điểm MN (gt)

\(\Rightarrow EM=EN=\dfrac{1}{2}MN\\ \Rightarrow EM=EN=6cm\)

Mà \(F\) là trung điểm \(EN\)

\(\RightarrowÈF=FN=\dfrac{1}{2}EN\\ \RightarrowÈ=FN=3cm\) 

Vậy...

Do E là trung điểm của MN

=> NE = MN : 2

= 12 : 2

= 6 (cm)

Do F là trung điểm của NE

=> EF = NE : 2

= 6 : 2

= 3 (cm)

\(1+\dfrac{3}{2}+2+\dfrac{5}{2}+...+\dfrac{199}{2}+100\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{200}{2}\)

\(=\dfrac{2+3+4+...+200}{2}\)

\(=\dfrac{199\cdot\dfrac{\left(200+2\right)}{2}}{2}=199\cdot\dfrac{202}{4}=10049,5\)

\(1+\dfrac{3}{2}+2+\dfrac{5}{2}+...+\dfrac{199}{2}+100\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{199}{2}+\dfrac{200}{2}\)

\(=\dfrac{2+3+4+5+...+199+200}{2}\)

Số số hạng của \(2+3+4+5+..+199+200\) là :

\(\left(200-2\right):1+1=199\)

Tổng là :

\(\left(200+2\right)\times199:2=20099\)

Ta có :

\(\dfrac{20099}{2}\) \(=10049,5\)

Lời giải:

Ta thấy, với $a,b,c,d>0$ thì:

$\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}> \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=\frac{a+b+c+d}{a+b+c+d}=1(*)$

Lại có:

Xét $\frac{a}{a+b+c}-\frac{a+d}{a+b+c+d}=\frac{-d(b+c)}{(a+b+c)(a+b+c+d)}<0$ với mọi $a,b,c,d>0$

$\Rightarrow \frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}$

Tương tự: $\frac{b}{b+c+d}< \frac{b+a}{a+b+c+d}, \frac{c}{c+d+a}< \frac{c+b}{a+b+c+d}; \frac{d}{d+a+b}< \frac{d+c}{a+b+c+d}$

Cộng lại suy ra:

$\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+d+b+a+c+b+d+c}{a+b+c+d}=\frac{2(a+b+c+d)}{a+b+c+d}=2(**)$

Từ $(*); (**)\Rightarrow$ đpcm.

11: \(2^2\cdot3^2-5\cdot2\cdot3=6^2-30=36-30=6\)

12: \(3^2\cdot5-2^2\cdot7+1\cdot5=9\cdot5-4\cdot7+5\)

=45-28+5

=50-28=22

13: \(5^2\cdot2-3^2\cdot4=25\cdot2-9\cdot4=50-36=14\)

14: \(7^2\cdot3-5^2\cdot3=49\cdot3-25\cdot3=24\cdot3=72\)

15: \(2^3\cdot3^2-4^2\cdot3=8\cdot9-16\cdot3=72-48=24\)

16: \(5^2\cdot2^3+3^2\cdot7-8^2\cdot2\)

\(=25\cdot8+9\cdot7-64\cdot2\)

=200+63-128

=263-128=135

17: \(\left(5\cdot2^2-20\right):5+3^2\cdot6=\left(5\cdot4-20\right):5+9\cdot6\)

=0+54

=54

18: \(\left(24\cdot5-5^2\cdot2\right):\left(5\cdot2\right)-3\)

\(=\left(120-50\right):10-3\)

=7-3=4

19: \(\left[\left(5^2\cdot2^3-7^2\cdot2\right):2\right]\cdot6-7\cdot2^5\)

\(=\left[5^2\cdot2^2-7^2\right]\cdot6-7\cdot32\)

=(100-49)*6-224

=51*6-224

=82

20: \(\left(6\cdot5^2-13\cdot7\right)\cdot2-2^3\left(7+3\right)\)

\(=\left(6\cdot25-91\right)\cdot2-8\cdot10\)

\(=\left(150-91\right)\cdot2-80\)

=118-80=38

$a(b-c)-a(b+d)$

$=ab-ac-ab-ad$

$=-ac-ad$

\(a\left(b-c\right)-a\left(b+d\right)\\ =a\left[\left(b-c\right)-\left(b+d\right)\right]\\ =a\left(b-c-b+d\right)\\ =a\left(d-c\right)\)

\(6\left(x\cdot3\right)\)

\(=6\cdot3x\)

\(=18x\)

=73.8-73.59 - 59.8 + 59.73

=584 - 4307 - 472 + 4307

=-7323 + 3835

=-3488

$73.(8-59)-59.(8-73)$

$=73.8-73.59-59.8+59.73$

$=73.8-59.8$

$=8.(73-59)$

$=8.14=112$