Giải các pt chứa ẩn ở mẫu sau:
a, \(\frac{12}{8+x^3}=1+\frac{1}{x+2}\)
b,\(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
c,\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)
a.\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
b.\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
c.\(\frac{12}{8-x^3}=1+\frac{1}{x+2}\)
d.\(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
e.\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)
\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)
\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)
Vô nghiệm
Giải các pt chứa ẩn ở mẫu:
a) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
b) \(\frac{2x-5}{x+5}=3\)
c) \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
a, Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
\(Đkxđ:\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
\(Pt\Leftrightarrow x\left(x+2\right)-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tmđk\right)\end{matrix}\right.\)
Vậy .........
\(b,Đkxđ:x\ne-5\)
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow x=20\left(tmđk\right)\)
Vậy .........
c, \(Đkxđ:x\ne3\)
Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktmđk\right)\end{matrix}\right.\)
Vậy ............
Giải các phương trình sau :
á) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
b) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\)tu giai ra de ma
Giải các phương trình:
\(a,\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(b,\frac{2}{4-x^2}+\frac{1}{x^2-2x}=\frac{x-4}{x^2+2x}\)
Giai pt
- \(\frac{^{\left(x+2\right)^2}}{2x-3}-1=\frac{x^2+10}{2x-3}\)
- \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
giải các pt sau
a)5X(X-2020)+X=2020
b)4(X-5)2-(2X+1)2=0
c)\(\frac{3X}{5}-\frac{2X+1}{3}=2-\frac{X-3}{15}\)
d)5X3+10X2+5X=0
e)2X3-8X=0
f)\(\frac{X^2+5}{25-X^2}=\frac{3}{X+5}+\frac{X}{X-5}\)
g)\(\frac{4}{2X-3}-\frac{4X}{9-4X^2}=\frac{1}{2X+3}\)
h)|2X-4|-15=1
i)20-3|2X+1|=17
k)|4X+2|-1,5=1
GIẢI GIÚP MÌNH NHANH VỚI NHA
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
\(a,5x\left(x-2020\right)+x=2020\)
\(< =>5x\left(x-2020\right)+x-2020=0\)
\(< =>\left(5x+1\right)\left(x-2020\right)=0\)
\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)
\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)
\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)
\(< =>4x^2-80x+100-4x^2-4x-1=0\)
\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
Giải phương trình chứa ẩn ở mẫu
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
b) \(\frac{3x}{x^2+x+1}+\frac{8x}{x^2+2x+1}+\frac{x}{x^2+3x+1}=\frac{16}{5}\)
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:
\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)
\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)
\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)
\(\Leftrightarrow2t^2+t-1=6t^2-6t\)
\(\Leftrightarrow-4t^2+7t-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)
Vậy phương trình vô nghiệm.