\(\left\{{}\begin{matrix}x\ne-\dfrac{5}{3}\\-\dfrac{2}{3x+5}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{5}{3}\\3x+5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{5}{3}\\x< -\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow x< -\dfrac{5}{3}\)

Ta có: \(x+y=7\Rightarrow\left(x+y\right)^2=49\Rightarrow x^2+y^2+2xy=49\)

Mà: \(x^2+y^2=25\Rightarrow2xy=24\Rightarrow xy=12\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7\left(25-12\right)=91\)

(Vì\(x+y=7;x^2+y^2=25;xy=12\))

\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)

\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)

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

\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

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

==>Sai đề không mem

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)

\(=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}\)

\(=\left(\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x+5\right)\left(x-5\right)}\right).\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)

\(=\frac{\left(x+x-5\right)\left(x-x+5\right)}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)

\(=\frac{5\left(2x-5\right)}{\left(x-5\right)}.\frac{1}{2x-5}+\frac{x}{5-x}\)

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

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

\(=\frac{-\left(x-5\right)}{x-5}\)

\(=-1\)

=> biểu thức  P k phụ thuộc vào x

ủa cái này hình như bn lạc lõn gì đấy trả lời r mà ?

\(\Leftrightarrow80\left(2x-9\right)=240\times39\)

\(\Leftrightarrow160x-720=9360\)

\(\Leftrightarrow160x=9360+720\)

\(\Leftrightarrow160x=10080\)

\(\Leftrightarrow x=63\)

MSC (mẫu số chung): 240

Quy đồng mẫu 2 vế và khử mẫu ta được:

2x - 9 = 117

     2x = 117 + 9 = 126

       x = 126 : 2 = 63

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)