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\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+3}=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^3=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\dfrac{1}{27}=\dfrac{28}{243}\\ \Rightarrow\dfrac{28}{27}\left(\dfrac{1}{3}\right)^x=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\dfrac{28}{243}:\dfrac{28}{27}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\dfrac{1}{9}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^2\\ \Rightarrow x=2\)
(1/3)x +(1/3)x+3=28/243 (1/3)x +(1/3)x *(1/3)3=28/243 (1/3)x *(1+1/33)=28/243 (1/3)x *28/27=28/243 (1/3)x=1/9 (1/3)x=(1/3)2 vậy x =2
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^2=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\left[1+\left(\dfrac{1}{3}\right)^2\right]=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\dfrac{10}{9}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{10}{243}:\dfrac{10}{9}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{27}\)
\(\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x=3\)
Lời giải:
a.
$(\frac{-1}{3})^3.x=\frac{1}{81}=(\frac{-1}{3})^4$
$\Rightarrow x=(\frac{-1}{3})^4: (\frac{-1}{3})^3=\frac{-1}{3}$
b.
$2^2.16> 2^x> 4^2$
$\Rightarrow 2^2.2^4> 2^x> (2^2)^2$
$\Rightarrow 2^6> 2^x> 2^4$
$\Rightarrow 6> x> 4$
$\Rightarrow x=5$ (với điều kiện $x$ là số tự nhiên nhé)
c.
$9.27< 3^x< 243$
$3.3^3< 3^x< 3^5$
$\Rightarrow 3^4< 3^x< 3^5$
$\Rightarrow 4< x< 5$
Với $x$ là stn thì không có số nào thỏa mãn.
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
a: =>x=(-2/3)^5:(-2/3)^2=(-2/3)^3=-8/27
b: =>x*(-1/3)^3=(-1/3)^4
=>x=-1/3
d: =>3x-2=-3
=>3x=-1
=>x=-1/3
a) Ta có: \(\left(4x-1\right)^2=\left(1-4x\right)^2\)
\(\Leftrightarrow\left(4x-1\right)^2-\left(1-4x\right)^2=0\)
\(\Leftrightarrow\left(4x-1-1+4x\right)\left(4x-1+1-4x\right)=0\)
\(\Leftrightarrow0\cdot x=0\)(luôn đúng)
Vậy: \(x\in R\)
b) Ta có: \(\dfrac{x-100}{24}+\dfrac{x-98}{26}+\dfrac{x-96}{28}=3\)
\(\Leftrightarrow\dfrac{x-100}{24}-1+\dfrac{x-98}{26}-1+\dfrac{x-96}{28}-1=0\)
\(\Leftrightarrow\dfrac{x-124}{24}+\dfrac{x-124}{26}+\dfrac{x-124}{28}=0\)
\(\Leftrightarrow\left(x-124\right)\cdot\left(\dfrac{1}{24}+\dfrac{1}{26}+\dfrac{1}{28}\right)=0\)
mà \(\dfrac{1}{24}+\dfrac{1}{16}+\dfrac{1}{28}>0\)
nên x-124=0
hay x=124
Vậy: x=124
a: =>x-1/2=1/3
=>x=5/6
b: =>|2x-1|=x+1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x-1-x-1\right)\left(2x-1+x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x-2\right)\left(3x\right)=0\end{matrix}\right.\)
hay \(x\in\left\{2;0\right\}\)
c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^x\cdot\dfrac{28}{27}=\dfrac{28}{243}\)
hay x=2
Bn giải chi tiết hơn hộ mik đc ko ạ