Ta có:\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{2z^2}{32}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)

\(\dfrac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=4\)

\(\dfrac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow y=6\)

\(\dfrac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow z=8\)

A=(\(\dfrac{1}{3}-\dfrac{1}{3}\))\(+\left(\dfrac{3}{5}+\left(\dfrac{-3}{5}\right)\right)+\left(\dfrac{-5}{7}+\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)\)\(+\left(\dfrac{-11}{13}-\dfrac{9}{11}\right)\)

A\(=0+0+0+0+\dfrac{-238}{143}\)

A\(=\dfrac{-238}{143}\)

\(B=\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{4}\right)+\left(1+\dfrac{1}{8}\right)+\left(1+\dfrac{1}{32}\right)+\left(1+\dfrac{1}{64}\right)-7\)

\(B=\left(1+1+1+1+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)-7\)

\(B=6+\dfrac{63}{64}-7\)

\(B=-1+\dfrac{63}{64}\)

\(B=\dfrac{-1}{64}\)

Ta có

\(A=2+2^2+2^3+...+2^{2010}=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)\(A=2.3+2^3.3+...+2^{2009}.3=\left(2+2^3+2^5+...+2^{2009}\right).3\)chia hết cho 3

\(5,2\in Q\) ;\(4,6351..........\in I\) ;\(-7,0903...\notin Q\) ;\(1,333\notin I\)

Tính tay á

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\)

\(\Rightarrow\)\(x=2k;y=5k;z=7k\)(1)

Thay (1) vào A ,ta có:

A=\(\dfrac{x-y+z}{x+2y-z}=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)

Vậy A=\(\dfrac{4}{5}\)

\(-\left|-2,15\right|=-2,15\)

a)Vì \(\dfrac{21}{52}>0\); \(\dfrac{-213}{523}< 0\)

\(\Rightarrow\)\(\dfrac{21}{52}>\dfrac{-213}{523}\)

b)Ta có :\(\dfrac{22}{35}< 1\) ;\(\dfrac{103}{17}>1\)

\(\Rightarrow\)\(\dfrac{22}{35}< \dfrac{103}{17}\)

c)Ta có :\(\dfrac{-2525}{4949}=\dfrac{-25}{49}\) ;\(\dfrac{-131313}{373737}=\dfrac{-13}{37}\)

Lại có :\(\dfrac{25}{49}>\dfrac{25}{50}=\dfrac{1}{2}\) hay \(\dfrac{-25}{49}< \dfrac{-25}{50}=\dfrac{-1}{2}\)(1)

và \(\dfrac{13}{37}< \dfrac{13}{26}=\dfrac{1}{2}\) hay \(\dfrac{-13}{37}>\dfrac{-13}{26}=\dfrac{-1}{2}\) (2)

Từ (1) và (2) \(\Rightarrow\)\(\dfrac{-2525}{4949}< \dfrac{-131313}{373737}\)