Question

tìm x

a, \(1\dfrac{2}{3}x-\dfrac{1}{4}=\dfrac{5}{6}\)

b, \(\dfrac{x}{-4}=\dfrac{-48}{3x}\)

c, \(3^{2.x+1}=81\)

Answer 1

b) x/-4=-48/3x

=>x.3x=-4.(-48)

=>3x²= 192

=>x²=64

=>x=8

Vậy............

Answer 2

\(a,1\dfrac{2}{3}x-\dfrac{1}{4}=\dfrac{5}{6}.\)

\(1\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{4}.\)

\(1\dfrac{2}{3}x=\dfrac{13}{12}.\)

\(x=\dfrac{13}{12}:1\dfrac{2}{3}.\)

\(x=\dfrac{13}{20}.\)

Vậy \(x=\dfrac{13}{20}.\)

\(c,3^{2x+1}=81.\)

\(3^{2x+1}=3^4.\)

\(\Rightarrow2x+1=4.\)

\(\Rightarrow2x=5.\)

\(\Rightarrow x=\dfrac{5}{2}.\)

Vậy \(x=\dfrac{5}{2}.\)

Answer 3

c) 3^2x+1=81

=>3^2x-1=3⁴

=>2x-1=4

=>x=5/2

Vậy....

Answer 4

c, $3^{2.x+1}=81$

$\mathrm{=>\; 32x+1=34}$

$\mathrm{=>\; 2x+1=4}$

$\mathrm{=>\; 2x=5}$

$\mathrm{=>\; x=\backslash (\backslash dfrac\{5\}\{2\}\backslash )}$