Giúp mik vs!!
Tìm x biết
2x+3 - 3.2x+1 = 32
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a: =>2x^3=58-4=54
=>x^3=27
=>x=3
b; =>(5-x)^5=2^5
=>5-x=2
=>x=3
c: =>(5x-6)^3=4^3
=>5x-6=4
=>5x=10
=>x=2
d: (3x)^3=(2x+1)^3
=>3x=2x+1
=>x=1
1=>2x3=54
=>x3=27 =>x=3
2=>(5-x)5=25
=>5-x=2
=>x=3
3=>(5x-6)3=43
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1
\(2x-10,01=19,01-3\\ \Rightarrow2x-10,01=16,01\\ \Rightarrow2x=16,01+10,01\\ \Leftrightarrow2x=26,02\\ \Leftrightarrow x=26,02:2=13,01\)
2x-10,01=19,01-3
=> 2x-10,01= 16,01
=> 2x= 16,01+10,01
=>2x= 26,02
=> x= 26,02: 2= 13,01
\(\Leftrightarrow2^x\cdot8-3\cdot2^x\cdot2=32\)
=>x+1=5
hay x=4
( x + 1 )5 = - 32
( x + 1 )5 = - 25
=> x + 1 = - 2
=> x = -3
Study well
\(x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{9}{14}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{4}{7}\)
\(x=\dfrac{19}{21}\)
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)
\(x=\dfrac{-2}{5}\)
Đặt A(x)=0
=>-x(-2x+3)(1-x^3)=0
=>x(2x-3)(x^3-1)=0
=>x=0 hoặc 2x-3=0 hoặc x^3-1=0
=>x=0;x=3/2;x=1
\(2^{x+3}-3.2^{x+1}=32\)
\(\Leftrightarrow2^x.2^3-3.2^x.2=32\)
\(\Leftrightarrow2^x.\left(2^3-3.2\right)=32\)
\(\Leftrightarrow2^x=16=2^4\)
\(\Leftrightarrow x=4\).
Cảm ơn bn!