5x - 25 = 10
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![](https://rs.olm.vn/images/avt/0.png?1311)
=(10x-x^2)(5x-25)
=50x^2-250x-5x^3+25x^2
=-5x^3+75x^2-250x
`@` `\text {Ans}`
`\downarrow`
`x(10-x)(5x-25)`
`= (10x-x^2)(5x-25)`
`= 10x(5x-25) - x^2(5x-25)`
`= 50x^2-250x - 5x^3 + 25x^2`
`= -5x^3+75x^2-250x`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(175-5\times\left(x-10\right)=25\)
\(\Rightarrow5\times\left(x-10\right)=175-25\)
\(\Rightarrow5\times\left(x-10\right)=150\)
\(\Rightarrow x-10=150:5\)
\(\Rightarrow x-10=30\)
\(\Rightarrow x=30+10\)
\(\Rightarrow x=40\)
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!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)
ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)
\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)
\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)
\(=\frac{77x+175}{15x^2+150x+375}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(x\ne\pm5\)
\(C=\dfrac{\left(x+2\right)\left(x-2\right)}{x^2-25}.\dfrac{x^2-25}{x^2+10}=\dfrac{x^2-4}{x^2+10}\)
\(C=2\Leftrightarrow x^2-4=2x^2+20\Leftrightarrow x^2=-24\left(vô-lí\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\Leftrightarrow x+5< 0\Leftrightarrow x< -5\)
5x - 25 = 10
5x = 10 + 25
5x = 35
x =35 : 5
x = 7
Vậy x = 7
5x - 25 = 10
5x = 10 + 25
5x = 35
x =35 : 5
x = 7
Vậy x = 7