\(\frac{3}{4}+\frac{3}{16}+\frac{3}{16} \)
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\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\Rightarrow\frac{1}{x}=\frac{11}{12}\Rightarrow x=\frac{12}{11}\)
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
=\(\frac{12}{16}+\frac{3}{16}+\frac{3}{16}\)
=\(\frac{18}{16}=\frac{9}{8}\)
\(c1:\frac{3}{4}+\frac{3}{16}+\frac{3}{16}=\frac{15}{16}+\frac{3}{16}=\frac{9}{8}.\)
\(c2:\frac{3}{4}+\frac{3}{16}+\frac{3}{16}=\frac{3}{4}+\frac{6}{16}=\frac{9}{8}\)