Tìm x
x^3 = -27/343
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\(x^3=-\frac{27}{343}\)
\(x^3=\left(-\frac{3}{7}\right)^3\)
\(x=-\frac{3}{7}\)
Bài làm:
Ta có: \(\sqrt{3}x-\sqrt{27}=\sqrt{343}\)
\(\Leftrightarrow\left(x-3\right)\sqrt{3}=7\sqrt{7}\)
\(\Leftrightarrow x-3=\frac{7\sqrt{21}}{3}\)
\(\Rightarrow x=\frac{9+7\sqrt{21}}{3}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(=\left(\frac{1}{2}:4\right):\frac{919191}{808080}=\frac{1}{8}\cdot\frac{808080}{919191}=\frac{10}{91}\)
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
a/ => (n - 1)3 = 23 => n - 1 = 2 => n = 3
b/ => n + 1 = 3 => n = 2
hoặc n + 1 = -3 => n = -4
c/ => (2n + 1)3 = 73 => 2n + 1= 7 => 2n = 6 => n = 3
d/ => (n - 2)3 = 33=> n - 2 = 3 => n =5
a,
(n-1)^3=8
(n-1)^3=2^3
=>n-1=2
n=2+1
n=3
b,(n+1)^2=9
(n+1)^2=3^2
=>n+1=3
n=3-2
n=1
tương tự như mấy câu trên
c,n=6
d,n=5
=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)
=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)
=\(182.\frac{1}{8}.\frac{91}{80}\)
=.\(182.\frac{91}{640}\)
=\(\frac{8281}{320}\)
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)