a) \(\left(2-\frac{3}{2}\right)\left(2-\frac{4}{3}\right)\left(2-\frac{5}{4}\right)\left(2-\frac{6}{4}\right)\)

\(=\frac{1}{3}\left(-\frac{4}{3}+2\right)\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)

\(=\frac{1}{2}.\frac{2}{3}\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\left(-\frac{6}{4}+2\right)\)

\(=\frac{1.2.3\left(2-\frac{3}{2}\right)}{2.3.4}\)

\(=\frac{1.3\left(2-\frac{3}{2}\right)}{3.4}\)

\(=\frac{1.\left(2-\frac{3}{2}\right)}{4}\)

\(=\frac{2-\frac{3}{4}}{4}\)

\(=\frac{1}{2.4}\)

\(=\frac{1}{8}\)

b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)

\(=\frac{8028024\left(\frac{2003}{2004}+\frac{2004}{2003}\right)}{8028025}\)

\(=\frac{8028024.\frac{8028025}{4014012}}{8028025}\)

\(=\frac{16056050}{8028025}\)

= 2

a) \(\frac{-77}{143}+\frac{65}{143}-\frac{66}{143}+\frac{7}{22}\)

= \(\frac{-78}{143}+\frac{7}{22}\)= \(\frac{-6}{11}+\frac{7}{22}\)= \(\frac{-12}{22}+\frac{7}{22}\)

= \(\frac{-5}{22}\)

b) \(\frac{-4}{5}-\frac{20}{170}+\frac{51}{170}+\frac{150}{170}\)= \(\frac{-4}{5}-\frac{221}{170}\)

\(\frac{-4}{5}-\frac{13}{10}\)= \(\frac{-8}{10}-\frac{13}{10}\)=\(\frac{-21}{10}\)

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)

\(4B=\frac{44}{45}\)

\(B=\frac{11}{45}\)

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Rightarrow2\cdot\frac{x-1}{2x+2}=\frac{2009}{2011}\)

\(\Rightarrow\frac{2x-2}{2x+2}=\frac{2009}{2011}\)

Bạn làm nốt.Nhân chéo là ra

\(\left(x-1\right)f\left(x\right)=\left(x+4\right)\cdot f\left(x+8\right)\)

Với  \(x=1\) ta có:

\(\left(1-1\right)\cdot f\left(1\right)=\left(1+4\right)\cdot f\left(9\right)\)

\(\Rightarrow5\cdot f\left(9\right)=0\)

\(\Rightarrow f\left(9\right)=0\)

Vậy \(x=9\)

Thay \(x=-4\) vào ta được:

\(\left(-4-1\right)\cdot f\left(-4\right)=0\cdot f\left(4\right)\)

\(\Rightarrow f\left(-4\right)=0\)

Vậy \(x=-4\)

\(\Rightarrow f\left(x\right)\) có ít nhất 2 nghiệm là 9;-4

c )

\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{5}{3}}}=1+\frac{1}{1+\frac{1}{\frac{8}{3}}}=1+\frac{1}{\frac{11}{8}}=\frac{19}{11}\)

1.

Theo bài ra ta có:

\(\frac{x}{2}=\frac{y}{3},\frac{y}{4}=\frac{z}{5}\) và x + y - z = 10

Ta có:

\(\frac{x}{8}=\frac{y}{12},\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

Suy ra:

x = 2 . 8 = 16

y = 2 . 12 = 24

z = 2 . 15 = 30

2/

Đặt \(\frac{x}{2}=\frac{y}{5}=k\)

Ta có :x = 2k ; y = 5k

=>x . y = 2k . 5k = 10k² = 10 => k² = 1 => k = ±1

Thay k = 1 ta có : x = 2 . 1 = 2     ;      y = 5 . 1 = 5

Thay k = -1 ta có : x = 2 . (-1) = -2    ;    y = 5 . (-1) = -5

Vậy x = ±2   ;  y = ±5

3/

Giải:

Gọi số học sinh khối 6,7,8,9 lần lượt là a,b,c,d .

Theo bài ra ta có:

\(\frac{a}{9}=\frac{b}{8}=\frac{c}{7}=\frac{d}{6}\) và b - d = 70

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{a}{9}=\frac{b}{8}=\frac{c}{7}=\frac{d}{6}=\frac{b-d}{8-6}=\frac{70}{2}=35\)

Suy ra :

a = 35 . 9 = 315

b = 35 . 8 = 280

c = 35 . 7 = 245

d = 35 . 6 = 210

Vậy số học sinh khối 6,7,8,9 lần lượt là 315;280;245;210 .