3/2 + ( x/6 + 1/3 )^2
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\(y\times3+\dfrac{y}{2}+\dfrac{y}{4}=1\dfrac{1}{2}\\ \Rightarrow y\times3+y\times\dfrac{1}{2}+y\times\dfrac{1}{4}=\dfrac{3}{2}\\ \Rightarrow y\times\left(3+\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{3}{2}\\ \Rightarrow y\times\dfrac{15}{4}=\dfrac{3}{2}\\ \Rightarrow y=\dfrac{3}{2}:\dfrac{15}{4}\\ \Rightarrow y=\dfrac{2}{5}\)
1: Ta có: \(S_1=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2020\right)+2021\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)
\(=-1\cdot1010+2021\)
\(=-1010+2021=1011\)
2) Ta có: \(S_2=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)
\(=\left(-2+4\right)+\left(-6+8\right)+...+\left(-2014+2016\right)\)
\(=2+2+...+2\)
\(=2\cdot504=1008\)
\(\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right).\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)
=> x = 1
Ta có: \(\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\cdot\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{1}{2}\)
hay x=1
Ta có: \(\dfrac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)^2}{4\cdot\left(\sqrt{5}+1\right)}\)
\(=\dfrac{4}{4}=1\)