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\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
Ta có \(1^2+2^2+3^2+...+49^2=\frac{49\left(49+1\right)\left(2.49+1\right)}{6}=40425\)
\(\Rightarrow40425\left(2-x\right)=-1\frac{1}{5}\Rightarrow2-x=-\frac{2}{67375}\Rightarrow x=\frac{134752}{67375}\)
a,(x-7)x+1-(x-7)x+11=0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0
=>x-7=0
=>x=7
hoặc 1-(x-7)10=1
=>(x-7)10=1
=>x-7=-1;1
=>x=8;6
vậy x=6;7;8
b,(x-1)2=36/49
=>x-1=6/7;-6/7
=>x=13/7;1/7
vậy x=1/7;13/7
a,(x-7)x+1-(x-7)x+11=0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0
=>x-7=0
=>x=7
hoặc 1-(x-7)10=1
=>(x-7)10=1
=>x-7=-1;1
=>x=8;6
vậy x=6;7;8
b,(x-1)2=36/49
=>x-1=6/7;-6/7
=>x=13/7;1/7
vậy x=1/7;13/7
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
Xét \(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(\frac{-31}{12}\right)^2=\frac{2401}{144}+\frac{961}{144}\)
\(\Rightarrow2x=\frac{3362}{144}\)
\(\Rightarrow x=\frac{3362}{144}.\frac{1}{2}=\frac{1681}{144}\)
Ta lai xét :
\(x+\left(\frac{-31}{12}\right)^2=y^2\)
\(\Rightarrow\frac{1681}{144}+\frac{-961}{144}=y^2\)
\(\Rightarrow\frac{720}{144}=y^2\)
\(\Rightarrow y^2=5\)
\(\Rightarrow y=2,236067977\)
\(\left(x-1\right)^2=49\)
\(\left(x-1\right)^2=\left(\pm7\right)^2\)
\(\Rightarrow\begin{cases}x-1=7\\x-1=-7\end{cases}\)
\(\Rightarrow\begin{cases}x=7+1\\x=-7+1\end{cases}\)
\(\Rightarrow\begin{cases}x=8\\x=-6\end{cases}\)
Vậy \(x\in\left\{8;-6\right\}\)
Ta có: (x-1)\(^2\) = 49
(x-1) = \(\left(\pm7\right)^2\)
* x-1 = 7 => x = 7+1=8.
* x-1= -7 => x= -7+1 = -6.
Vậy x=8 hoặc x= -6.
\(\left(x-1\right)^2=49\)
=> \(\left(x-1\right)^2=7^2=\left(-7\right)^2\)
=> \(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=> \(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
KL: x thuộc {8; -6}
x=8 nha