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a.
\(-x-\dfrac{2}{3}=\dfrac{-6}{7}\)
\(-x=\dfrac{-6}{7}+\dfrac{2}{3}\)
\(-x=\dfrac{-4}{21}\)
\(\Rightarrow x=\dfrac{4}{21}\)
Vậy\(x=\dfrac{4}{21}\)
b.
\(x^2=16\)
\(x^2=4^2\)
\(\Rightarrow x=4\)
Vậy x = 4
c.
\(2^{x-1}=64\)
\(2^{x-1}=2^6\)
\(\Rightarrow x-1=6\)
\(x=6+1\)
\(x=7\)
Vậy x = 7
Giải:
a) \(-x-\dfrac{2}{3}=-\dfrac{6}{7}\)
\(\Leftrightarrow-x=-\dfrac{6}{7}+\dfrac{2}{3}\)
\(\Leftrightarrow-x=-\dfrac{4}{21}\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
b) \(x^2=16\)
\(x^2=4^2\)
Vì \(2=2\)
Nên \(x=4\)
c) \(2^{x-1}=64\)
\(2^{x-1}=2^6\)
Vì \(2=2\)
Nên \(x-1=6\)
\(\Leftrightarrow x=6+1\)
\(\Leftrightarrow x=7\)
Chúc bạn học tốt!
a) \(3^{-8}\le3^x\le\frac{1}{243}.\)
\(\Rightarrow3^{-8}\le3^x\le3^{-5}\)
\(\Rightarrow-8\le x\le-5\)
\(\Rightarrow\left[{}\begin{matrix}x=-8\\x=-7\\x=-6\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{-8;-7;-6;-5\right\}.\)
b) \(2^{-x}< \frac{1}{64}\)
\(\Rightarrow2^{-x}< 2^{-6}\)
\(\Rightarrow-x< -6\)
\(\Rightarrow x< 6.\)
Vậy \(x< 6.\)
Chúc bạn học tốt!
\(\left|x\right|=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}\)
b,\(\left(2x-3\right)^2=64\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm8\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=8\\2x-3=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{5}{2}\end{cases}}}\)
c,\(\left(\frac{1}{2}\right)^x=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{2}\right)^x=\left(\pm\frac{1}{2}\right)^4\)
\(\Rightarrow x=\pm\frac{1}{2}\)
d,\(3^{x+1}=27\)
\(\Leftrightarrow3^{x+1}=3^3\)
\(\Leftrightarrow x+1=3\)
\(\Leftrightarrow x=2\)
b) (x-1)x+2 =(x-1)2
=> x+2=2
=> x=0
vậy
c) (1-3x)3=-64
=> 1-3x=-4
=> -3x=-4-1
=> -3x=-5
=> x=5/3
vậy ....
Làm nốt ::v
\(a.\left(x+2\right)^2=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow\left(x+2\right)^2=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{\dfrac{1}{6}}\\x+2=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{1}{6}}-2\\x=-\sqrt{\dfrac{1}{6}}-2\end{matrix}\right.\)
KL...........
a/ \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\Leftrightarrow\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\Leftrightarrow\dfrac{52}{3x}=20\)
\(\Leftrightarrow x=\dfrac{13}{15}\)
Vậy..
b/ \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy..
c/ \(\left(2^3:4\right).2^{x+1}=64\)
\(\Leftrightarrow2.2^{x+1}=64\)
\(\Leftrightarrow2^{x+2}=2^6\)
\(\Leftrightarrow x+2=6\)
\(\Leftrightarrow x=4\)
Vậy..
d/ \(\left|3-2x\right|-3=-3\)
\(\Leftrightarrow\left|3-2x\right|=0\)
\(\Leftrightarrow3-2x=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy..
e/ \(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\)
\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)
Vậy..
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