\(\dfrac{97}{96}\)và \(\dfrac{199}{197}\)

\(\dfrac{97}{96}=\dfrac{19109}{18912}\),\(\dfrac{199}{197}=\dfrac{19104}{18912}\)

\(\Rightarrow\dfrac{19109}{18912}>\dfrac{19102}{18912}\)

vậy \(\dfrac{97}{96}>\dfrac{199}{197}\)

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

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

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

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

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

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

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

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

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

$=73.8-73.59-59.8+59.73$

$=73.8-59.8$

$=8.(73-59)$

$=8.14=112$

8= 2³ ; 27 = 3³ ; 125= 5³ ; 216 = 6³ ; 343 = 7³ ; 8000 = 20³

\(14\cdot98=14\cdot\left(100-2\right)=14\cdot100-14\cdot2\)

=1400-28

=1372

cảm ơn

Câu 1:

a: B

b: D

c: D

d: 12;15;18;a

a là trung bình cộng của 4 số nên ta có: \(a=\dfrac{12+15+18+a}{4}\)

=>\(4a=a+45\)

=>3a=45

=>a=15

=>Chọn A

Câu 2:

Số tiền lãi là:

\(500000\cdot20\%=100000\left(đồng\right)\)

Câu 3:

Trong 1 giờ, anh Thành làm được: \(\dfrac{1}{12}\)(công việc)

Trong 1 giờ, anh Công làm được: \(\dfrac{1}{12:2}=\dfrac{1}{6}\)(công việc)

Trong 1 giờ, hai anh làm được: \(\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{1}{12}+\dfrac{2}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)(công việc)

=>Hai anh cần 1:1/4=4 giờ để hoàn thành công việc khi làm chung