bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

nhớ làm giúp mình nhá mai mình phải đi học r :<

a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)

\(x-\frac{7}{9}=\frac{4}{9}\)

\(x=\frac{4}{9}+\frac{7}{9}\)

\(x=\frac{11}{9}\)

Vậy x=\(\frac{11}{9}\)

\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)

\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{99}{100}-\frac{99}{100}\)

\(A=\frac{99-99}{100}=0\)

Bài 2 

\(\left(3x+5\right).\left(2x-4\right)=0\)

\(TH1:3x+5=0\)

\(3x=-5\)

\(x=-\frac{5}{3}\)

\(TH2:2x-4=0\)

\(2x=4\)

\(x=2\)

\(\left(x^2-1\right).\left(x+3\right)=0\)

\(\Rightarrow x^2-1=0\)

\(x^2=1\)

\(\Rightarrow x=1\)

\(x+3=0\)

\(x=-3\)

\(5x^2-\frac{1}{2}x=0\)

\(\Rightarrow5x^2-\frac{x}{2}=0\)

\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)

\(10x^2-x=x.\left(10x-1\right)\)

\(\frac{x.\left(10x-1\right)}{2}=0\)

\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)

\(10x-1=0\)

\(x=\frac{1}{10}=0.100\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)

\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)

\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)

\(\frac{x}{4}=\frac{5}{4}\)

\(\Rightarrow x=5\)

\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)

\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)

\(x=\frac{7}{8}:\frac{5}{8}\)

\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)

\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(A=\frac{99}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{99}{100}-\frac{99}{100}\)

\(=0\)

Câu A

X + (X+1) + (X+3) +...+ (X+2003) = 2004 

Số số hạng trong tổng 1 + 3 + ... + 2003 là

(2003 - 1) : 2 + 1 = 1002

Tổng dãy 1 + 3 + ... + 2003 là:

(1 + 2003) * 1002 : 2 = 1004004

=> (1003.X) + 1004004 = 2004

=>                  (1003.X)= 2004 - 1004004

=>                  1003.X = - 1002000

                        X = - 1002000/1003

E chỉ giải đc đến đây thui!!!!!!!!!!!!!!! :)))

x + ( x + 1) + (x + 3) ... + (x + 2003) = 2004

x + x + x + ... + x (có 1003 x) + 1 + 3 + 5 + ... + 2003 = 2004

x . 1003 + 1004004 = 2004

x . 1003 = 2004 - 1004004

x . 1003 = -1002000

x = -1002000 : 1003

x = -999,00299 = ~-999

a,Khai triển biểu thức ra ta được:

1003x+1004004=2004\(\Leftrightarrow\)1003x=-1002000\(\Leftrightarrow\)x=\(\frac{-1002000}{1003}\)

b,\(\left(x+2\right)^2=\frac{1}{6}\Leftrightarrow\orbr{\begin{cases}x+2=\frac{1}{\sqrt{6}}\\x+2=-\frac{1}{\sqrt{6}}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{\sqrt{6}}-2\\x=-\frac{1}{\sqrt{6}}-2\end{cases}}}\)

c,\(\left(2x+1\right)^2=25\Leftrightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

d,Cộng 3 vào 2 vế ta có:

\(\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1+\frac{x-11}{12}+1\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

Vì \(\hept{\begin{cases}\frac{1}{7}>\frac{1}{10}\\\frac{1}{8}>\frac{1}{11}\\\frac{1}{9}>\frac{1}{12}\end{cases}\Rightarrow\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}>0\Rightarrow x+1=0\Leftrightarrow x=-1}\)