THUC HIEN PHEP TINH:\(\frac{^{x^2}}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}\)
LAM ON GIUP MINH VS LAM HOAI MA NO CU SAI T.T
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ND
1
HN
3 tháng 12 2018
Ta có :\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x}\)
\(=\frac{4\left(x-2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{2\left(x+2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{\left(5x-6\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{4x-2x^2+16}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32+4x-2x^2+16+5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{32x-5x^2-28}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
PN
1
ND
11 tháng 12 2017
\(a.\)
\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
\(=\left(x-5\right)\left(x+5\right).\dfrac{3x-7}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
\(b.\)
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\dfrac{5\left(x-1\right)}{3\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}\)
\(=\dfrac{x}{3\left(x-1\right)}\)
NH
11 tháng 12 2017
\(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}=\dfrac{5x\left(x+1\right)\left(x-1\right)}{15\left(x-1\right)^2\left(x+1\right)}=\dfrac{x}{3\left(x-1\right)}\)\(\left(x^2-25\right):\dfrac{2x+10}{3x-7}=\dfrac{\left(x-5\right)\left(x+5\right)\left(3x-7\right)}{2\left(x+5\right)}=\dfrac{\left(x-5\right)\left(3x-7\right)}{2}\)
30 tháng 7 2018
a/ \(\left|5x+\frac{3}{4}\right|-\frac{5}{4}=2\)
\(\left|5x+\frac{3}{4}\right|=\frac{13}{4}\)
- / \(5x+\frac{3}{4}=\frac{13}{4}\)
\(5x=\frac{5}{2}\)
\(x=\frac{1}{2}\) - / \(5x+\frac{3}{4}=-\frac{13}{4}\)
\(5x=-4\)
\(x=-\frac{4}{5}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{4}{5}\right\}\)
b/\(\frac{3}{2}-\left|\frac{1}{2}x+1\right|=\frac{1}{4}\)
\(\left|\frac{1}{2}x+1\right|=\frac{5}{4}\)
1/\(\frac{1}{2}x+1=\frac{5}{4}\)
\(\frac{1}{2}x=\frac{1}{4}\)
\(x=\frac{1}{2}\)
2/\(\frac{1}{2}x+1=-\frac{5}{4}\)
\(\frac{1}{2}x=-\frac{9}{4}\)
\(x=-\frac{9}{2}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{9}{2}\right\}\)
MT
1
12 tháng 10 2020
\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)
ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)
\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)
\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)
\(=\frac{77x+175}{15x^2+150x+375}\)
PN
1
TH
1
DH
15 tháng 2 2020
\(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)
\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\left(ktmđk\right)\)
Vậy pt vô nghiệm
\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)