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a)Ta có:
\(3^x-3^{x-3}=-234\)
\(\Rightarrow3^x-3^x\cdot3^3=-234\)
\(\Rightarrow3^x\cdot\left(1-3^3\right)=-234\)
\(\Rightarrow3^x\cdot\left(-26\right)=-234\)
\(\Rightarrow3^x=9\)
\(\Rightarrow x=2\)
Vậy x=2
\(\Rightarrow3^x=3^2\)
b) Ta có:
\(2^{x+1}\cdot3^x-6^x=216\)
\(\Rightarrow2^x\cdot2\cdot3^x-2^x\cdot3^x=216\)
\(\Rightarrow\left(2^x\cdot3^x\right)\cdot\left(2-1\right)=216\)
\(\Rightarrow6^x\cdot1=216\)
\(\Rightarrow6^x=6^3\)
\(\Rightarrow x=3\)
Vậy x=3
a. 6,5 -9/4:/x+1/3\=/-2\
6,5-9/4:/x+1/3\=2
9/4:/x+1/3\=6,5-2
9/4:/x+1/3\=4,5
/x+1/3\=9/4:4,5
/x+1/3\=1/2
x+1/3=1/2 hoặc x+1/3= -1/2
x= 1/2-1/3 x= -1/2-1/3
x= 1/6 x= -5/6
Vậy x=1/6 hoặcx= -5/6
b. 2-/3/2x-1/4\ = /-5/4\
2-/3/2x-1/4\=5/4
/3/2x-1/4\=2-5/4
/3/2x-1/4\=3/4
3/2x-1/4=3/4 hoặc 3/2x-1/4= -3/4
3/2x=3/4+1/4 3/2x= -3/4+1/4
3/2x=1 3/2x= -1/2
x=1:3/2 x= -1/2:3/2
x=2/3 x= -1/3
Vậy x=2/3 hoặc x= -1/3
/x/3+1/=2/3
(=)x/3+1=2/3 hoặc -2/3
Trường hợp 1: x/3+1=2/3
=) x/3= 2/3-1
=) x/3=-1/3
=)X=-1
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
a) \(1,25-\left|0,5-x\right|=0\)
\(\Rightarrow\left|0,5-x\right|=1,25\)
\(\Rightarrow\begin{cases}0,5-x=1,25\\0,5-x=-1,25\end{cases}\)
\(\Rightarrow\begin{cases}x=-0,75\\x=1,75\end{cases}\)
Vậy \(x=\begin{cases}-0,75\\1,75\end{cases}\)
a) \(1,25-\left|0,5-x\right|=0\)
\(\Leftrightarrow\left|0,5-x\right|=1,25\)
+)TH1: x =< 0,5 thì pt trở thành:
0,5-x=1,25 <=> x=-0,75 (tm)
+)TH2: x>0,5 thì pt trở thành:
x-0,5=1,25 <=> x=1,75 (tm)
Vậy x={-0,75;1,75}
b) \(\left|2x\right|-\left|-3,5\right|=\left|-6,5\right|\)
\(\Leftrightarrow\left|2x\right|=10\)
+)Th1: x>= 0 thì pt trở thành:
2x=10 <=> x=5 (tm)
+)TH2: x<0 thì pt trở thành:
2x=-10 <=> x=-5 (tm)
Vậy x={-5;5}
c) \(\left|2x+1\right|+1=4\Leftrightarrow\left|2x+1\right|=3\)
+)TH1: x>= -1/2 thì pt trở thành:
2x+1=3 <=> x=1 (tm)
+)TH2: x<-1/2 thì pt trở thành:
2x+1=-3 <=> 2x=-4 <=> x=-2 (tm)
Vậy x={-2;1}
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
b) Ta có: \(2^{2x+1}+4^{x+3}=264\)
\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)
\(\Leftrightarrow4^x=4\)
hay x=1