\(1\dfrac{1}{2}+1\dfrac{1}{4}+1\dfrac{1}{8}+1\dfrac{1}{16}+1\dfrac{1}{32}+1\dfrac{1}{64}\)

\(=1+\dfrac{1}{2}+1+\dfrac{1}{4}+1+\dfrac{1}{8}+1+\dfrac{1}{16}+1+\dfrac{1}{32}+1+\dfrac{1}{64}\)

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

\(=6+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(=6+\left(1-\dfrac{1}{64}\right)=6+\dfrac{63}{64}=\dfrac{447}{64}\)

\(2x\left(1-2x\right)+\left(x-1\right)^2=2x-4x^2+x^2-2x+1=-3x^2+1=\left(1-\sqrt{3}x\right)\left(1+\sqrt{3}x\right)\)

Gọi số hsg 3 lớp 7A 7B 7C lần lượt là x,y,z (học sinh; x,y,z \(\in\)N*)

Có số hsg 3 lớp tỉ lệ với 2;4;6 ta có x:2=y:4=z:6

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

Số hsg lớp 7C nhiều hơn 7B là 6 em => z-x=6

Áp dụng tc dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{z-x}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=9\end{matrix}\right.\) (tm)

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