bằng còn cái nịt

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2021}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right)....\left(-\frac{2020}{2021}\right)\)

\(=\frac{1.2.3...2020}{2.3.4...2021}=\frac{1}{2021}\)

Ta có

(1−1/2)×(1−1/3)×(1−1/4).....×(1−1/2020)×(1−1/2021)(1-1/2)×(1-1/3)×(1-1/4).....×(1-1/2020)×(1-1/2021)

=1/2×2/3×3/4.....×2019/2020×2020/2021=1/2×2/3×3/4.....×2019/2020×2020/2021

=1×2×3×.....×2019×2020/2×3×4×....×2020×2021=1×2×3×.....×2019×2020/2×3×4×....×2020×2021

=1/2021

 \(N=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+..+\frac{10}{1993.2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)=\frac{3}{10}.\frac{2000}{6009}=\frac{200}{2003}\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{3}{13.23}\)\(+\)\(\frac{3}{23.33}\)\(+...+\)\(\frac{3}{1993.2003}\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}.\frac{1990}{26039}\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{597}{26039}\)

\(N=\)\(\frac{200}{2003}\)

Trả lời:

Làm tròn các số sau đến hàng chục, trăm, nghìn, chục nghìn:   

a) 763189\(\approx\)763190

b) 198575\(\approx\)198600

c) 2398761\(\approx\)2399000

d) 1895678\(\approx\)1900000

HT

ối dồi ôi bạn ơi, toán lớp 1 chứ không phải lớp 7, tôi xốk quá

4)

Ta có x/3=y/2 và x/4=z/5

=>x/12=y/8=z/15

Theo tình chất các tỉ số bằng nhau ta có 

X/12=y/8=z/15=x+y-z/12+8-15=10/5=2

=>x=2.12=24

 y=8.2=16

 z=15.2=30

Kết luận:.......

bài nào bn?

a. ta có :

\(\hept{\begin{cases}\left|x-1\right|+\left|x-4\right|\ge\left|x-1-x+4\right|=3\\\left|x-2\right|+\left|x-3\right|\ge\left|x-2-x+3\right|=1\\\left|2x-5\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\)thay lại thấy thỏa mãn . Vậy x=5/2 là nghiệm

b.ta có 

\(\hept{\begin{cases}\left|x+1\right|+\left|x-1\right|\ge\left|x+1-x+1\right|=2\\\left|x+2\right|+\left|x-5\right|\ge\left|x+2-x+5\right|=7\\\left|3x+2\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\)thay lại thấy thỏa mãn . Vậy x=-2/3 là nghiệm

Ta có \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

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

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

Khi a + b + c + d => a + b = -(c + d) ; 

b + c = -(a + d) ; 

c + d = -(a + b) 

d + a = -(b + c)

Khi đó \(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)

= (-1) + (-1) + (-1) + (-1) = -4

Khi a + b + c + d \(\ne0\)

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)

Khi đó M = \(\frac{2a}{2a}+\frac{2b}{2b}+\frac{2c}{2c}+\frac{2d}{2d}=2+2+2+2=8\)

Vậy khi a + b + c + d = 0 thì M = -4

khi a + b + c + d \(\ne\)0 thì M = 8

thanks bn nha!!!!!!!!!!!!

ok luôn bài này là max nhé :

\(A=-3,7-\left|1,7-m\right|\le-3,7\forall m\)  (vì : \(\left|1,7-m\right|\ge0\forall m\) )

" = " <=> m = 1,7

Vậy ... 

giải giùm tui

TH1: \(x\le-1\)

ta có phương trình \(\left|x+1\right|+\left|2x-5\right|+\left|x-9\right|=10\Leftrightarrow-x-1-2x+5-x+9=10\)

\(\Leftrightarrow-4x=-3\Leftrightarrow x=\frac{3}{4}\left(\text{loại}\right)\)

TH2: \(-1< x\le\frac{5}{2}\) thì

\(\left|x+1\right|+\left|2x-5\right|+\left|x-9\right|=10\Leftrightarrow x+1-2x+5-x+9=10\)

\(\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\left(tm\right)\)

Th3: \(\frac{5}{2}< x\le9\) thì

\(\left|x+1\right|+\left|2x-5\right|+\left|x-9\right|=10\Leftrightarrow x+1+2x-5-x+9=10\)

\(\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\left(\text{loại}\right)\)

th4:\(x>9\)thì 

\(\left|x+1\right|+\left|2x-5\right|+\left|x-9\right|=10\Leftrightarrow x+1+2x-5+x-9=10\)

\(\Leftrightarrow4x=23\Leftrightarrow x=\frac{23}{4}\left(\text{loại}\right)\) 

Vậy x=5/2