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lop 6 thi cha chua hoc con gi nua 

moi cap 1 con doi

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\Rightarrow\frac{1}{x}=\frac{11}{12}\Rightarrow x=\frac{12}{11}\)

hoặc \(\frac{1}{x}-\frac{2}{3}=-\frac{1}{4}\Rightarrow\frac{1}{x}=\frac{5}{12}\Rightarrow x=\frac{12}{5}\)

Vậy x = 12/11 , x = 12/5

Theo giả thiết, ta có:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow\)  \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Rightarrow\)  \(2\left(ab+bc+ac\right)=0\)

\(\Rightarrow\)  \(ab+bc+ac=0\)

Vì   \(a,b,c\ne0\)  nên  \(\frac{ab+bc+ac}{abc}=0\), tức là  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)  \(\left(1\right)\)

Từ  \(\left(1\right)\)  \(\Rightarrow\)  \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)  \(\left(2\right)\)

               \(\Leftrightarrow\)   \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

               \(\Leftrightarrow\)  \(\frac{1}{a^3}+\frac{1}{b^3}+3.\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

               \(\Leftrightarrow\)  \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)

               \(\Leftrightarrow\)   \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)  (do  \(\left(2\right)\) )

\(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{31}.3^{16}}=\frac{2^{50}.3^{16}+3^{45}+2^{50}+2^{40}.3^{16}}{2^{31}+3^{20}+2^{31}.3^{16}}\)

                                 \(=556758,4881\)

=\(\frac{2^{50}.+2^{90}}{2^{51}+2^{31}}=\frac{2^{19}}{2^{39}}=\frac{1}{1048576}\)

K nha