(x-3)^6-1/1/3=62/1/3
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NT
0
6 tháng 1 2022
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
6 tháng 1 2022
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
NN
1
12 tháng 7 2021
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
Q
21 tháng 4 2017
a, \(\dfrac{62}{7}.x=\dfrac{29}{90}.\dfrac{3}{56}\)
\(\dfrac{62}{7}.x=\dfrac{29}{1680}\)
\(x=\dfrac{29}{1680}:\dfrac{62}{7}\)
\(x=\dfrac{29}{14880}\)
b, \(\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)
\(\dfrac{1}{5}:x=\dfrac{2}{35}\)
\(x=\dfrac{1}{5}:\dfrac{2}{35}\)
\(x=\dfrac{7}{2}\)
c, \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{13}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(\left(x+\dfrac{-1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\)
\(\left(x+\dfrac{-1}{12}\right)=\dfrac{7}{46}.\dfrac{23}{12}\)
\(x+\dfrac{-1}{12}=\dfrac{7}{24}\)
\(x=\dfrac{7}{24}-\dfrac{-1}{12}\)
\(x=\dfrac{3}{8}\)
\(\left(x-3\right)^6-\frac{1}{\frac{1}{3}}=\frac{62}{\frac{1}{3}}\)
\(\left(x-3\right)^6=\frac{62}{\frac{1}{3}}+\frac{1}{\frac{1}{3}}\)
\(\left(x-3\right)^6=62:\frac{1}{3}+1:\frac{1}{3}\)
\(\left(x-3\right)^6=\left(62+1\right):\frac{1}{3}\)
\(\left(x-3\right)^6=63.\frac{1}{3}\)
\(\left(x-3\right)^6=\frac{63}{3}\)
\(\left(x-3\right)^6=21\)
\(\Rightarrow x-3\in\varnothing\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)