1, Tìm x
a, 2*3^x=3^12*34+20*27^4
b, (2^x+1)^2+3*(2^2+1)=2^2*10
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1: Tìm x
a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)
\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)
\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)
\(\Leftrightarrow3^x=3^{12}\cdot3^3\)
\(\Leftrightarrow3^x=3^{15}\)
hay x=15
Vậy: x=15
b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)
\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)
Vậy: x=2
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
a,3/4 . (-5/12)+3/4.(-7/12)
` 3/4 . [ - ( 5/12 + 7/12 ) ] `
`3/4 . (-1) = -3/4 `
`2/3 . x - 0,5 = 3/4 `
` x - 0,5 = 3/4 - 2/3 `
` x-0,5 = 1/12 `
` x = 1/12 + 0,5 `
` x= 7/12 `
Bài 2:
a: =>2/3x=3/4+1/2=3/4+2/4=5/4
=>x=5/4:2/3=5/4*3/2=15/8
b:=>-2x+4=3x-12
=>-5x=-16
=>x=16/5
11: Ta có: \(\left(x+3\right)^3=125\)
\(\Leftrightarrow x+3=5\)
hay x=2
12: Ta có: \(\left(2x\right)^4=16\)
\(\Leftrightarrow x^4=1\)
hay \(x\in\left\{1;-1\right\}\)