Ta có : \(\frac{4^{x+2}+4^{x+1}+4^x}{21}=\frac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}\)

\(\Rightarrow\frac{4^x\left(4^2+4+1\right)}{21}=\frac{3^{2x}\left(1+3+3^3\right)}{31}\)

\(\Rightarrow\frac{4^x.21}{21}=\frac{3^{2x}.31}{31}\)

=> 4^x = 3^2x

=> 4^x = (3²)^x

=> 4^x = 9^x

=> \(\frac{4^x}{9^x}=1\)(vì lũy thừa của một số khác 0 luôn luôn là 1 số khác 0)

=> \(\left(\frac{4}{9}\right)^x=1\)

=> x = 0 

Vậy x = 0

= \(\dfrac{2x^3+5-x^3-4}{x^2-x+1}\) = \(\dfrac{x^3-1}{x^2-x+1}\)

\(\dfrac{2x^3+5 -x^3-4}{x^2-x+1}=\dfrac{x^3+1 }{x+1}\)

câu này luôn \(\dfrac{2x^2y^5}{xy^2z}-\dfrac{4x^2y^3}{xy^2z}\)

vì \(x^4+2x^2+1=\left(x^2+1\right)^2\) mà \(x^2\ge0\Rightarrow x^2+1>0\Rightarrow\left(x^2+1\right)^2>0\)với mọi x.Nên x-3=0 .Từ đó suy ra x=3

\(\left(2x-\dfrac{3}{4}\right)^2=\left(3-x\right)^2\)

\(\Rightarrow2x-\dfrac{3}{4}=3-x\)

\(3x=3\dfrac{3}{4}\)

\(x=\dfrac{5}{4}\)

a) (2x - 1) x - x (2x + 3) = 7

<=> x (2x - 1 - 2x - 3) = 7

<=> -4x = 7

<=> x = \(-\dfrac{7}{4}\)

b) 3 (2x - 1) - 5 (x - 3) + 6 (3x - 4) = 24

<=> 6x - 3 - 5x + 15 + 18x - 24 = 24

<=> 19x - 12 = 24

<=> 19x = 36

<=> x = \(\dfrac{36}{19}\)

Đặt \(\sqrt[3]{x^2+1}=t\left(t\ge1\right)\)

\(y=f\left(t\right)=t^2-t+1\)

\(minf\left(t\right)=f\left(1\right)=1\)

\(minf\left(t\right)=1\Leftrightarrow t=1\Leftrightarrow\sqrt[3]{x^2+1}=1\Leftrightarrow x=0\)

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)