\(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(5x-1\right)\left(3-5x\right)}\)
Những câu hỏi liên quan
\(\frac{3}{5x-1}-\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(ĐKXĐ:x\ne\frac{1}{5},x\ne\frac{3}{5}\)
Ta có : \(\frac{3}{5x-1}=\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\frac{3\left(3-5x\right)}{\left(5x-1\right)\left(3-5x\right)}-\frac{2\left(5x-1\right)}{\left(5x-1\right)\left(3-5x\right)}+\frac{4}{\left(5x-1\right)\left(5x-3\right)}=0\)
\(\Rightarrow9-15x-10x+2+4=0\)
\(\Leftrightarrow-25x=-15\)
\(\Leftrightarrow x=\frac{3}{5}\) ( không thỏa mãn \(ĐKXĐ\) )
Vậy pt đã cho vô nghiệm
26 ,giải phương trình.
a,\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
b,\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
c,\(\frac{x-1}{x+2}+\frac{x-2}{x+1}=\frac{2\left(x^2+2\right)}{x^2-4}\)
d,\(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
https://i.imgur.com/V92CPVX.jpg
https://i.imgur.com/yXrzVbQ.jpg
\(\frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2+x+x^2-3x=4x\Leftrightarrow2x^2-6x=0\Leftrightarrow2x\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
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Tìm x biết
\(a,-\frac{1}{2}\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
\(b,\sqrt{9\left(5x-1\right)}-\sqrt{16\cdot\left(5x-1\right)}+\sqrt{36\left(5x-1\right)}=15\)
MÌNH ĐANG CẦN GẤP GIẢI CỤ THỂ GIÚP MÌNH NHA
Giai Phuong Trinh :
\(\frac{3}{5x-1}\)+\(\frac{2}{3-5x}\)=\(\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\frac{-3x.\left(5x+3\right)}{1+3x}>=0\)\(\frac{-2x^2+5x-3}{-x.\left(3x+7\right)}>0\)\(\frac{1}{x-2}-\frac{4}{x^2-4}< \frac{1}{3}\)\(x^2-20x+51>0\)\(\left(x-3\right).\left(2x+1\right)\left(1-5x\right)< 0\)\(\left(x-2\right)\left(x+3\right)=< 0\)
Giải Phương trình sau
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
b) \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
3x+2/3x-2 - 6/2+3x = 9x2/9x2-4 (đkxđ :3/2 , -3/2)
<=>3x+2/(3x-2).(3x+2) - 6/(3x-2).(3x+2) = 9x2/(3x-2).(3x+2)
=>3x+2 - 6 = 9x2
<=>3x-9x2=6+2
<=>3x-9x2=8
<=>3(x-3x)=8
<=>x-3x=8/3
<=>2x=8/3
<=>x=8/3 / 2
<=>x=4/3(thoải mãn)
vậy phương trình có nghiệm x = 4/3
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bây giờ phải đi học thêm rồi phần b làm cũng tương tự phần a mà cố gắng lên nha
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1)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
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Giải các phương trình sau:
a) left(x-3right)left(1-xright)left(2x-3right)left(x+5right)
b) frac{3}{5x-1}+frac{2}{3-5x}frac{4}{left(1-5xright)left(x-3right)}
c) frac{x+1}{x^2+x+1}-frac{x-1}{x^2-x-1}frac{3}{xleft(x^4+x^2+1right)}
d) frac{m-3}{m+3}-frac{1-3m}{1+3m}-1
e)frac{x}{3+x}-frac{x}{x-3}frac{1}{9-x^2}
Đọc tiếp
Giải các phương trình sau:
a) \(\left(x-3\right)\left(1-x\right)=\left(2x-3\right)\left(x+5\right)\)
b) \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
c) \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x-1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
d) \(\frac{m-3}{m+3}-\frac{1-3m}{1+3m}-1\)
e)\(\frac{x}{3+x}-\frac{x}{x-3}=\frac{1}{9-x^2}\)
Bài 1: Tìm x biết:
a, \(x.\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
b, \(\left(5x-1\right).\left(2x-\frac{1}{3}\right)=0\)
c, \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
d, \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
e, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-1\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
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e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
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