ĐKXĐ : \(x\ne\pm5\)

\(C=\dfrac{\left(x+2\right)\left(x-2\right)}{x^2-25}.\dfrac{x^2-25}{x^2+10}=\dfrac{x^2-4}{x^2+10}\)

\(C=2\Leftrightarrow x^2-4=2x^2+20\Leftrightarrow x^2=-24\left(vô-lí\right)\)

\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)

ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)

\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)

\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)

\(=\frac{77x+175}{15x^2+150x+375}\)

\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

d, ĐKXĐ:\(x\ne-2,x\ne3\)

\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)

a/ => -2x + 3x = 25 - 10 

=> x = 15

b/=> 4x - 3 - 5x = 14 

=> -x = 17 => x = -17

|a|<=b khi -b<=a<=b

Tui làm nốt câu cuối thui,2 câu đầu ông làm rùi

a)\(P=\dfrac{x^2}{5x+25}+\dfrac{2x-10}{x}+\dfrac{50+5x}{x^2+5x}\left(đkxđ:x\ne0;-5\right)\)

\(P=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(P=\dfrac{x^3+10\left(x+5\right)\left(x-5\right)+5\left(50+5x\right)}{5x\left(x+5\right)}\)

\(P=\dfrac{x^3+10x^2-250+250+25x}{5x\left(x+5\right)}\)

\(P=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}\)

\(P=\dfrac{x+5}{5}\)

b)P âm\(\Leftrightarrow x+5< 0\Leftrightarrow x< -5\)

\(175-5\times\left(x-10\right)=25\)

\(\Rightarrow5\times\left(x-10\right)=175-25\)

\(\Rightarrow5\times\left(x-10\right)=150\)

\(\Rightarrow x-10=150:5\)

\(\Rightarrow x-10=30\)

\(\Rightarrow x=30+10\)

\(\Rightarrow x=40\)

Học tốt #

\(175-5\left(x-10\right)=25\)

\(\Leftrightarrow5\left(x-10\right)=150\)

\(\Leftrightarrow x-10=30\)

\(\Leftrightarrow x=40\)

hok tốt nhé!

5 * (x- 10 ) = 25+ 175

5* (x- 10 ) = 200

x- 10 = 200/ 5

x- 10 = 40

x = 40 + 10

x =50

\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)

\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

a ) \(\left(x+4\right)^2+\left(1-x\right)\left(1+x\right)=7\)

\(\Leftrightarrow x^2+8x+16+1^2-x^2=7\)

\(\Leftrightarrow8x+17=7\)

\(\Leftrightarrow8x+10=0\)

\(\Rightarrow x=-\dfrac{10}{8}\)