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a) \(2^3.2^x=64\Leftrightarrow8.2^x=64\Leftrightarrow2^x=\dfrac{64}{8}=8=2^3\Rightarrow x=3\)
vậy \(x=3\)
b) \(7.7^x=343\Leftrightarrow7^x=\dfrac{343}{7}=49=7^2\Rightarrow x=2\)
vậy \(x=2\)
c) \(7.7^{x+1}=343\Leftrightarrow7^{x+1}=\dfrac{343}{7}=49=7^2\Rightarrow x+1=2\Leftrightarrow x=1\)
vậy \(x=1\)
d) \(2^x-15=17\Leftrightarrow2^x=17+15=32=2^5\Rightarrow x=5\)
vậy \(x=5\)
e) \(x^{10}=1\Leftrightarrow x^{10}=1^{10}\Rightarrow x=1\)
vậy \(x=1\)
\(x^{10}=x\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy \(x=0;x=1\)
a) \(2^3.2^x=2^3.2^3\)
=> \(2^x=2^3\)
=> x=3
b) \(7.7^x=343\)
=> \(7.7^x=7.7^2\)
=> \(7^x=7^2\)
=> x=2
c) \(7.7^{x+1}=7.7^2\)
=> \(7^{x+1}=7^2\)
=> x+1=2
=> x=1
d) \(2^x-15=17\)
=> \(2^x=17+15\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> x=5
e) \(x^{10}=1\)
=> \(x^{10}=1^{10}\)
=> x=1
g) \(x^{10}=x\)
=> x = 0 hoặc 1
+) Với x=0 => \(0^{10}=0\)( t/m )
+) Với x=1 => \(1^{10}=1\)( t/m )

a) 2x=64:23=64:8=8=23=>x=3
b) 7x=343:7=49=72=> x=2
c) Tương tự như câu trên, với x+1 thì x=1
d) 2x=17+15=32=25=>x=5
e) => \(x\in\left(-1;1\right)\)
g) =>x=0;1
Ta có : x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

a) \(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
b) \(2^x.16=128\)
\(2^x=128:16\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
c) \(3^x:9=27\)
\(3^x=27.9\)
\(3^x=243\)
\(3^x=3^5\)
\(\Rightarrow x=5\)
d) \(x^4=x\)
\(\Rightarrow x=0\)hoac \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
e) \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
f) \(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)
\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)
\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)
\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)
\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)
b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)
d) \(x^4=x\Leftrightarrow x=1\)
e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)
\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)
F)

1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4

320 : ( x - 1 ) = ( 53 - 52 ) : 4 + 15
<=> 320 : ( x - 1 ) = ( 125 - 25 ) : 4 + 15
<=> 320 : ( x - 1 ) = 100 : 4 + 15
<=> 320 : ( x - 1 ) = 40
<=> x - 1 = 8
<=> x = 9
240 : ( x - 5 ) = 22 . 52 - 20
<=> 240 : ( x - 5 ) = 4 . 25 - 20
<=> 240 : ( x - 5 ) = 80
<=> x - 5 = 3
<=> x = 8
70 : ( x - 3 ) = ( 34 - 1 ) : 4 - 16
<=> 70 : ( x - 3 ) = 80 : 4 - 16
<=> 70 : ( x - 3 ) = 4
<=> x - 3 = 35/2
<=> x = 41/2
a) 320: (x-1) = (53 - 52):4 + 15 b) 240: (x-5) = 22.52 - 20 c) 70:(x-3) = (34 - 1):4 - 16
320: (x-1) = (125- 25): 4 + 15 240: (x-5) = 4.25 - 20 70:(x-3) = (81-1): 4 - 16
320: (x-1) = 100:4 + 15 240: (x-5) = 100 - 20 70: (x-3) = 80: 4 - 16
320: (x-1) = 25 + 15 240: (x-5) = 80 70: (x-3) = 20 -16
320: (x-1) = 40 (x-5) = 240:80 70: (x-3) = 4
(x-1) = 320:40 (x-5) = 3 (x-3) = 70:4
(x-1) = 80 x = 5+3 (x-3) = 17,5
x = 80+1 x = 8 x = 17,5 : 3
x = 81 x = 5,83

a,5/10-2/3x=7/12 b, x:4 1/3=-2,5 c, 5,5x=13/15
2/3x=5/10-7/12 x:13/3=-25/10 55/10x=13/15
2/3x=30/60-35/60 x=-25/10.13/3 x=13/15:55/10
2/3x=-1/12 x=-65/6 x=13/15.10/55
x=-1/12:2/3 x=26/165
x=-1/12.3/2
x=-1/8
d,(3x/7+1):(-4)=-1/28
(3x/7+1) = -1/28.(-4)
(3x/7+1) =1/7
3x/7 =1/7-1
3x/7 =-6/7
=> 3x=-6
x=-6:3
x=-2
=>x=-2
\(\rArr\) \(\frac12-x=8\) hoặc \(\frac12-x=-8\)
+) Với \(\frac12-x=8\)
\(x=\) \(\frac12-8\)
\(x=-\frac{15}{2}\)
+) Với \(\frac12-x=-8\)
\(x=\frac12-\left(-8\right)\)
\(x=\frac{17}{2}\)
Vậy \(x=\frac{17}{2}\) hoặc \(x=-\frac{15}{2}\)
(1/2 - x)^2 = 64
(1/2 - x)^2 = 8 = -8
[còn lại làm theo 2 trường hợp]