Gọi 3 số hữu tỉ có dạng chung là x. Ta có: 

\(-\frac{6}{7}< x< \frac{1}{-3}\)

\(\Leftrightarrow\frac{-6}{7}< x< \frac{-1}{3}\)

\(\Leftrightarrow\frac{-18}{21}< x< \frac{-7}{21}\)

\(\Rightarrow x=\left\{\frac{-17}{21};\frac{-16}{21};\frac{-15}{21}\right\}\)

câu hỏi là gì ?!

vẽ hình theo d=đó

Ta có : \(x^2-4x+4=x^2-2x-2x+4=x\left(x-2\right)-2\left(x-2\right)=\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\)Có vô số x

Vậy có vô số x

Ta có x² - 4x + 4 \(\ge0\)(1)

<=> x² - 2x - 2x + 4  \(\ge0\)

<=> x(x - 2) - 2(x - 2)  \(\ge0\)

<=> (x - 2)(x - 2)  \(\ge0\)

<=> (x - 2)²  \(\ge0\)(đúng với mọi x)

Vậy \(\forall x\inℝ\)thì (1) đúng

c)\(-\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}+\frac{1}{1999.2000}\right)\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\right)\)

\(=-\left(1-\frac{1}{2000}\right)=-\frac{1999}{2000}\)

d)\(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)

\(=-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

\(=-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

\(=-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(=-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=-\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(=-\frac{1}{2}.\frac{100}{101}=-\frac{50}{101}\)

\(9^x=27^4.81\)

\(\Rightarrow\left(3^2\right)^x=\left(3^3\right)^4.3^4\)

\(\Rightarrow3^{2x}=3^{12}.3^4\)

\(\Rightarrow3^{2x}=3^{16}\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Vậy \(x=8\)

9 ^x = 27 ⁴.  81

( 3² ) ^x =  ( 3³ ) ⁴. 3⁴

3 ^2x = 3¹² . 3⁴

3 ^2x  = 3¹⁶

2 x = 16

   x = 8

Ta có: 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

Phương trình ban đầu tương đương với: 

\(2x-\frac{49}{50}=7-\frac{1}{50}+x\)

\(\Leftrightarrow x=7-\frac{1}{50}+\frac{49}{50}=\frac{198}{25}\)

Đề là thế này \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=17-\frac{1}{50+x}\) đúng không em ??

\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)

\(\Rightarrow x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=7-\frac{1}{50}\)

\(\Rightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=7-\frac{1}{50}\)

\(\Rightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=7-\frac{1}{50}\)

\(\Rightarrow x-\left(1-\frac{1}{50}\right)=7-\frac{1}{50}\)

\(\Rightarrow x=7-\frac{1}{50}+\left(1-\frac{1}{50}\right)\)

\(\Rightarrow x=8-\frac{1}{25}\)

\(\Rightarrow x=\frac{199}{25}\)

Vậy \(x=\frac{199}{25}\)