A)2^x-1=64 b)3^2x-10=81 c)2^2x+3=1024 d) 618-3^x+1=24.5^2
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a, 2x.16 = 1024 => 2x = 1024:16 => 2x = 64 => 2x = 26 => x = 6
b, x17 = x
=> x17 - x = 0
=> x(x16-1)=0
=> x = 0 hoặc x16 - 1 = 0
=> x = 0 hoặc x16 = 1
=> x = 0 hoặc x = 1
c, (2x-2)3=64
=> (2x-2)3 = 43
=>2x-2=4
=>2x=6
=>X=3
d,(x-6)2 = (x-6)3
=> (x-6)2-(x-6)3=0
=> (x-6)2-[1-(x-6)] = 0
=> (x-6)2 = 0 hoặc 1 - (x-6) = 0
=> x - 6 = 0 hoặc x - 6 = 1
=> x = 6 hoặc x = 7
e, 3 + 2x-1 = 24-[42-(22-1)]
=> 3 + 2x-1 = 11
=> 2x-1 = 8
=> 2x-1 = 23
=>x-1=3
=>x=4
a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
c: =>2x+1=4 hoặc 2x+1=-4
=>x=3/2 hoặc x=-5/2
h: =>x-3=4
=>x=7
g: =>2x-1=3
=>2x=4
=>x=2
f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)
=>x*-5/6=5/6
=>x=-1
d: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>x=-1 hoặc x=2
a: \(2^{x^2-1}=256\)
=>\(2^{x^2-1}=2^8\)
=>\(x^2-1=8\)
=>\(x^2=9\)
=>\(x\in\left\{3;-3\right\}\)
b: \(3^{x^2+3x}=81\)
=>\(3^{x^2+3x}=3^4\)
=>\(x^2+3x=4\)
=>\(x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
c: \(2^{x^2-5x}=64\)
=>\(2^{x^2-5x}=2^6\)
=>\(x^2-5x=6\)
=>\(x^2-5x-6=0\)
=>(x-6)(x+1)=0
=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^x=243\)
=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)
=>x=-5
e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)
=>\(3^{-x-5}=3^{2x+1}\)
=>-x-5=2x+1
=>-3x=6
=>x=-2
3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
a ) 9 . 3x = 81
3x = 9
x = 3
b ) 2x : 4 = 1
2x = 4
x = 2
c ) 2x - 64 = 2
2x = 66
x = 33
d ) 2x = 16
x = 8
e ) 3 ^ 2 . 3 ^ 4 . 3x = 3 ^ 10
3 ^ ( 2 + 4 + x ) = 3 ^ 10
=> 2 + 4 + x = 10
x = 4
f ) 2x + 4 . 2x = 5 . 2 ^ 5
2x + 8 x = 160
10x = 160
x = 16
a,
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)
\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
b,
( 3x + 1)^3 = 64
(3x + 1)^3 = 4^3
(3x + 1) = 4
3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
c,
( 2x - 3)^4 = 81
( 2x - 3) ^4 = 3^4
(2x - 3) = 3
2x = 3 + 3
2x = 6
x = 6: 2
x = 3
a, 2x-1 = 64
=> 2x-1 = 26
=> x - 1 = 6
=> x = 6 + 1
=> x = 7
b, 32x-10 = 81
=> 32x-10 = 34
=> 2x - 10 = 4
=> 2x = 14
=> x = 7
c, 22x+3 = 1024
=> 22x+3 = 210
=> 2x + 3 = 10
=> 2x = 7
=> x = \(\frac{7}{2}\)