\(5x.\left(3x-2\right)=4-9x^2\)

\(\Rightarrow5x.\left(3x-2\right)-\left(4-9x^2\right)=0\)

\(\Rightarrow5x.\left(3x-2\right)+\left(9x^2-4\right)=0\)

\(\Rightarrow5x.\left(3x-2\right)+\left(3x-2\right).\left(3x+2\right)=0\)

\(\Rightarrow\left(3x-2\right).\left(5x+3x+2\right)=0\)

\(\Rightarrow\left(3x-2\right).\left(8x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\8x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}\)

cảm ơn bạn

\(\left(x^2-y^2+6x+9\right):\left(x+y+3\right)\)

\(=[\left(x^2+2x.3+3^2\right)-y^2]:\left(x+y+3\right)\)

\(=[\left(x+3\right)^2-y^2]:\left(x+y+3\right)\)

\(=\left(x+3+y\right).\left(x+3-y\right):\left(x+y+3\right)\)

\(=\left(x+3-y\right)\)

x=3,605 nha 

x^3 - 64 = 0

x^3 = 0+64

x^3 = 64

Suy ra x^3 = 4^3 vì 64 = 4.4.4

Mà 4.4.4 = 4^3. Vậy x= 4

\(x^3-64=0\)

\(\left(x-4\right)\left(x^2+4x+16\right)=0\)

Dễ thấy \(x^2+4x+16=\left(x+2\right)^2+12>0\)

Nên x = 4

a) Điều kiện: \(x\ne\left\{0;\pm2\right\}\)

\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=[\frac{x^2}{x.\left(x-2\right).\left(x+2\right)}-\frac{6}{3.\left(x-2\right)}+\frac{1}{x+2}]:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2.\left(x+2\right)+x-2}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

\(=\frac{6}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

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

b) \(A\) \(Max\)

\(\Rightarrow-\frac{1}{x-2}Max\)

\(\Rightarrow\frac{1}{x-2}Min\)

\(\Rightarrow\left(x-2\right)\) \(Max\)

\(\Rightarrow x\) \(Max\)

\(\Rightarrow x\in\varnothing\)