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Ta có : \(\frac{5.5}{1.6}+\frac{5.5}{6.11}+\frac{5.5}{11.16}+\frac{5.5}{16.21}\)
\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)\)
\(=5\left(1-\frac{1}{21}\right)\)
\(=5.\frac{20}{21}=\frac{100}{21}\)
[5.5.5.5.5......5.5].[5.5.5.5-125.5]=A-B
B=5.5.5.5-125.5=5^4-5^3.5=5^4-5^4=0
A.0=0
`@` `\text {Ans}`
`\downarrow`
`(7x - 11)^2 = 2^5 * 5^2 + 100?`
`=> (7x - 11)^2 = 32*25 + 100`
`=> (7x - 11)^2 = 800 + 100`
`=> (7x - 11)^2 = 900`
`=> (7x - 11)^2 = (+-30)^2`
`=>`\(\left[{}\begin{matrix}7x-11=30\\7x-11=-30\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}7x=41\\7x=-19\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{41}{7}\\x=-\dfrac{19}{7}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{\dfrac{41}{7};-\dfrac{19}{7}\right\}\)
\(\left(x-1\right)\cdot\left(x-1\right)=5\cdot5\)
\(\left(x-1\right)^2=5^2\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5+1\\x=-5+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
(x - 1) . (x - 1) = 5,5
\(\left(x-1\right)^2\) = 5,5
x - 1 = \(\sqrt{5.5}\)
x = \(\sqrt{5.5}-1\)
50
5.5+5.5=25+25=50