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\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)
\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)
\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)
\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)
==>Sai đề không mem
=\(\left(\frac{x}{\left(x-5\right).\left(x+5\right)}-\frac{\left(x-5\right)}{x.\left(x+5\right)}\right).\frac{x^2+5x}{2x-5}\)
=\(\left(\frac{x^2}{x.\left(x-5\right).\left(x+5\right)}-\frac{\left(x-5\right)^2}{x.\left(x-5\right).\left(x+5\right)}\right).\frac{x\left(x+5\right)}{2x-5}\)
=\(\frac{x^2-\left(x-5\right)^2}{x.\left(x-5\right).\left(x+5\right)}.\frac{x.\left(x+5\right)}{2x-5}\)
=\(\frac{\left(x-x+5\right).\left(x+x-5\right)}{x.\left(x-5\right)\left(x+5\right)}.\frac{x.\left(x+5\right)}{2x+5}\)
=\(\frac{5.\left(2x-5\right).x\left(x+5\right)}{x.\left(x-5\right).\left(x+5\right).\left(2x-5\right)}\)
=\(\frac{5}{x+5}\)
P/s : lười làm nên đăng hình ảnh zậy , viết mỏi tay lắm ( em lùng ảnh cũ , ko phải bây h mới làm , có kí tên nên ko pải hàng fake )
\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)
ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)
\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)
\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)
\(=\frac{77x+175}{15x^2+150x+375}\)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
a, (2x+5)mũ 2=(x+2) mũ 2
=.> (2x+5) mũ 2-(x+2) mũ 2=0
=> (2x+5+x+2)x(2x+5-x-2)=0
=>(3x+7)x(x+3)=0
=>3x+7=0 hoặc x+3=0
3x+7=0=>x=-7/3
x+3=0 =>x=-3
vậy x=-7/3 hoặc x=-3
hok tot
ĐK: ...
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)( ko t/m )
d) tương tự, ngại tính lắm
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}-\frac{3x^2}{x^3-1}=\frac{2x\left(x-1\right)}{x^3-1}\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)
phân tích lần ra , rồi rút gọn
\(\left(\frac{3x-5}{x^2-5x}-\frac{x+5}{5x-25}\right):\frac{x^2-25}{x}\)
\(=\left[\frac{3x-5}{x\left(x-5\right)}-\frac{x+5}{5\left(x-5\right)}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{\left(3x-5\right).5}{x\left(x-5\right).5}-\frac{\left(x+5\right).x}{5\left(x-5\right).x}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{15x-25}{5x\left(x-5\right)}-\frac{x^2+5x}{5x\left(x-5\right)}\right].\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{15x-25-x^2-5x}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-x^2+10x-25}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-\left(x-5\right)^2.x}{5x\left(x-5\right)\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-1}{5\left(x+5\right)}\).