\(\Leftrightarrow\left(3x+7\right)\left(2x-5\right)=0\)

=>x=-7/3 hoặc x=5/2

\(2x\left(3x+7\right)-15x-35=0\\ \Rightarrow2x\left(3x+7\right)-\left(15x+35\right)=0\\ \Rightarrow2x\left(3x+7\right)-5\left(3x+7\right)=0\\ \Rightarrow\left(2x-5\right)\left(3x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(a,\left(x-2\right)^2-x\left(x+2\right)=20\\ \Leftrightarrow x^2-4x+4-x^2-2x=20\\ \Leftrightarrow-6x+4=20\\ \Leftrightarrow-6x=16\\ \Leftrightarrow x=-\dfrac{8}{3}\)

\(\Leftrightarrow x^2-4x+4-x^2-2x=20\)

=>-6x=16

hay x=-8/3

các pro giúp em vớiiiiiiiiii ;-;

\(\left(x^2-2x+3\right)\left(\frac{1}{2x}-5\right)\)

\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)

\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)

\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)

\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)

học tốt

(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15

a) Ta có: \(4\left(x-2\right)^2+xy-2y\)

\(=4\left(x-2\right)^2+y\left(x-2\right)\)

\(=\left(x-2\right)\left(4x-8+y\right)\)

b) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)

\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)

\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)

a) 2x - 3 = x + 1/2

<=> 2x - 3 = 1/2x + 1/2

<=> 2x - 3 - 1/2x = 1/2

<=> 3/2x - 3 = 1/2

<=> 3/2x = 1/2 + 3

<=> 3/2x = 7/2

<=> x = 7/2 : 3/2

<=> x = 7/3

=> x = 7/3

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

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

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

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

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

\(\frac{3}{2}x=\frac{7}{2}\)

\(x=\frac{7}{2}:\frac{3}{2}\)

\(x=\frac{14}{6}=\frac{7}{3}\)

\(\)B làm tương tự