Tính
1+2+3+....+100
1+3+5+....+101
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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: \(S_2=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)
\(=\left(-2+4\right)+\left(-6+8\right)+...+\left(-2014+2016\right)\)
\(=2+2+...+2\)
\(=2\cdot504=1008\)
1: \(\dfrac{12}{5\sqrt{6}}=\dfrac{12\sqrt{6}}{30}=\dfrac{2\sqrt{6}}{5}\)
2: \(\dfrac{3}{2+\sqrt{6}}=\dfrac{-6+3\sqrt{6}}{2}\)
\(\dfrac{1}{12}\times\dfrac{4}{5}=\dfrac{4}{60}=\dfrac{1}{15}\\ \dfrac{9}{5}:\dfrac{4}{7}=\dfrac{9}{5}\times\dfrac{7}{4}=\dfrac{63}{20}\\ 4\times\dfrac{3}{7}=\dfrac{4\times3}{7}=\dfrac{12}{7}\\ \dfrac{1}{2}:5=\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{1}{10}\)
13x204-204x3
= (13 - 3 ) x 204
= 10 x 204
= 2040
1001x101-1001
= 1001x101-1001x1
=1001x ( 101 - 1 )
=1001 x 100
=100100
k mik nha!
1: \(\left(x-1\right)^3-x\left(x-2\right)^2+1\)
\(=x^3-3x^2+3x-1-x^3+4x^2-4x+1\)
\(=x^2-x\)
2: \(2x\left(3x+2\right)-3x\left(2x+3\right)\)
\(=6x^2+4x-6x^2-9x\)
=-5x
\(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6}{30}-\dfrac{5}{30}=\dfrac{1}{30}\)
\(\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{7}{42}-\dfrac{6}{42}=\dfrac{1}{42}\)
`@mt`