\(\Leftrightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4.5\right)=3.5\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4.5=3.5\)

=>-6x=8

hay x=-4/3

\(2x-\left(x-3\right)=\frac{1}{2}\)

\(\Leftrightarrow2x-x+3=\frac{1}{2}\)

\(\Leftrightarrow2x-x=\frac{1}{2}-3\)

\(\Leftrightarrow x\left(2-1\right)=\frac{1}{2}-\frac{6}{2}\)

\(\Leftrightarrow x=\frac{-5}{2}\)

\(2x-\left(x-3\right)=\frac{1}{2}\)

\(\Leftrightarrow2x-x+3=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}-3\)

\(\Leftrightarrow x=\frac{1}{2}-\frac{6}{2}=-\frac{5}{2}\)

2x- x + 3 = 1/2

x = 1/2 - 3 

x=- 5 /2

a. \(2\left(x-\dfrac{1}{3}\right)+3x=\dfrac{7}{4}\)

=> \(2x-\dfrac{2}{3}+3x=\dfrac{7}{4}\)

=> \(2x+3x=\dfrac{7}{4}+\dfrac{2}{3}\)

=> \(5x=\dfrac{29}{12}\)

=> \(x=\dfrac{29}{60}\)

Câu b và c vế 2 mk ko biết dấu / của bạn là chia hay phân số

#Yiin

#)Giải :

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-5x-4,5=3,5\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\frac{8}{5}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-6x=8\)

\(\Leftrightarrow x=\frac{-8}{6}=\frac{-4}{3}\)

\(\Leftrightarrow\frac{7^x.7+7^x.7^2+7^x}{57}=\frac{5^{2x}.1+5^{2x}.5+5^{2x}.5^3}{131}\)

\(\Leftrightarrow\frac{7^x\left(7+49+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Leftrightarrow\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

<=> 7^x = 5^2x 

<=> \(\frac{7^x}{5^{2x}}=1\)

<=> \(\frac{7^x}{25^x}=1\)

<=> \(\left(\frac{7}{25}\right)^x=1\)

<=> x = 0