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PN
2
KB
28 tháng 3 2022
\(lim\dfrac{2\sqrt{7n^2-2n}}{3n+2}=lim\dfrac{2\sqrt{n^2\left(7-\dfrac{2}{n}\right)}}{3n+2}=lim\dfrac{2n\sqrt{7-\dfrac{2}{n}}}{n\left(3+\dfrac{2}{n}\right)}\)
\(=lim\dfrac{2\sqrt{7-\dfrac{2}{n}}}{3+\dfrac{2}{n}}=\dfrac{2\sqrt{7}}{3}\) \(=\dfrac{a\sqrt{7}}{b}\)
Suy ra : a/b = 2/3 => a - b = -1
L8
1
TT
0
28 tháng 6 2023
=>2/6+2/12+2/20+...+2/x(x+1)=1010/1011
=>2(1/6+1/12+...+1/x(x+1))=1010/1011
=>1/2-1/3+1/3-1/4+...+1/x-1/x+1=505/1011
=>1/x+1=1/2022
=>x+1=2022
=>x=2021
a, \(\left(x+y+x\right)^2-2\left(x+y+x\right)\left(y+z\right)+\left(y+z\right)^2\)
\(=\left(x+y+x-y-z\right)^2=\left(2x-z\right)^2\)
b, \(\left(x+y+x\right)^2-\left(y+z\right)^2=\left(2x+y\right)^2-\left(y+z\right)^2\)
\(=\left(2x+y-y-z\right)\left(2x+y+y+z\right)=\left(2x-z\right)\left(2x+2y+z\right)\)
c, \(\left(x+3\right)^2+4\left(x+3\right)+4=\left(x+3+2\right)^2=\left(x+5\right)^2\)
d, \(25+10\left(x+1\right)+\left(x+1\right)^2=\left(5+x+1\right)^2=\left(x+6\right)^2\)
e, \(\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2=\left(x+2+x-2\right)^2=\left(2x\right)^2\)
f, \(\left(x-3\right)^2-2\left(x^2-9\right)+\left(x+3\right)^2=\left(x-3\right)^2-2\left(x-3\right)\left(x+3\right)+\left(x+3\right)^2\)
\(=\left(x-3-x-3\right)^2=\left(-6\right)^2=6^2\)