\(5+\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
Những câu hỏi liên quan
Giải pt: \(\dfrac{3}{5x-1}+\dfrac{2}{3-5x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\dfrac{5+96}{x^2-16}=\dfrac{2x—1}{x+4}-\dfrac{3x-1}{4-x}\)
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
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a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
b) \(\dfrac{2x-1}{x+4}-\dfrac{1-3x}{x-4}=5+\dfrac{96}{x^2-16}\)
c) \(\dfrac{x+3}{x+1}-\dfrac{x-1}{x}=\dfrac{3x^2+4x+1}{x\left(x+1\right)}\)
5+\(\dfrac{96}{x^2-16}\)=\(\dfrac{2x-1}{x+4}\)-\(\dfrac{3x-1}{4-x}\)
\(5+\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\left(ĐKXĐ:x\ne\pm4\right)\)
\(\Leftrightarrow5+\dfrac{96}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{5\left(x-4\right)\left(x+4\right)+96}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
\(\Rightarrow5\left(x-4\right)\left(x+4\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
P/s: tới đây pn tự giải tiếp nha, kết quả \(x=8\) (nhận)
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Giải pt
a, \(\dfrac{5x-1}{3x+2}\)= \(\dfrac{5x-7}{3x-1}\)
b, \(\dfrac{4x+7}{x-1}\)= \(\dfrac{12x+5}{3x+4}\)
c, 5 + \(\dfrac{96}{x^2-16}\)=\(\dfrac{2x-1}{x+4}\)- \(\dfrac{3x+1}{4-x}\)
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
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giải phương trình
1)\(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
2)\(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
3)\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\)
4)\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}+\dfrac{3x-1}{x-4}-6\)
5)\(\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}-6\)
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
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1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
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3)
\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\\ \Leftrightarrow\dfrac{2\left(x-3\right)}{6}-\dfrac{3\left(x+2\right)}{6}=\dfrac{x}{6}\\ \Leftrightarrow2x-6-3x-6=x\\ \Leftrightarrow2x-3x-x=6+6\\ \Leftrightarrow-2x=12\\ \Leftrightarrow x=-6\)
Vậy PT có tập nghiệm S = { -6 }
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Giải các phương trình sau:
\(e.\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(f.\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
\(g.\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
\(h.\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
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Giải các phương trình sau :
1.dfrac{14}{3x-12}-dfrac{2+x}{x-4}dfrac{3}{8-2x}-dfrac{5}{6}
2.dfrac{12}{1-9x^2}dfrac{1-3x}{1+3x}-dfrac{1+3x}{1-3x}
3.dfrac{x+5}{x^2-5x}-dfrac{x+25}{2x^2-50}dfrac{x-5}{2x^2+10x}
4.dfrac{6x_{ }+1}{x^2-7x+10}+dfrac{5}{x-2}dfrac{3}{x-5}
5.dfrac{2}{x^2-4}-dfrac{x-1}{xleft(x-2right)}+dfrac{x-4}{xleft(x+2right)}0
6.dfrac{2}{x+2}-dfrac{2x^2+16}{x^3+8}dfrac{5}{x^2-2x+4}
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Giải các phương trình sau :
1.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
2.\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
3.\(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10x}\)
4.\(\dfrac{6x_{ }+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
5.\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)
6.\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
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a,2.(dfrac{1}{4}+x)^3(-dfrac{27}{4})b,(x+dfrac{1}{2})^3:3dfrac{-1}{81}c,(dfrac{2}{3}-x)^21:dfrac{4}{9}d,(2x-dfrac{1}{5})^2+dfrac{16}{25}1e,(dfrac{2}{5}-3x)^2-dfrac{1}{5}dfrac{4}{25}
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a,2.(\(\dfrac{1}{4}\)+x)\(^3\)=(\(-\dfrac{27}{4}\))
b,(x+\(\dfrac{1}{2}\))\(^3\):3=\(\dfrac{-1}{81}\)
c,(\(\dfrac{2}{3}\)-x)\(^2\)=1:\(\dfrac{4}{9}\)
d,(2x-\(\dfrac{1}{5}\))\(^2\)+\(\dfrac{16}{25}\)=1
e,(\(\dfrac{2}{5}\)-3x)\(^2\)-\(\dfrac{1}{5}\)=\(\dfrac{4}{25}\)
1.giải pt
a)\(\dfrac{16-x}{4}=\dfrac{2x+1}{3}\)
b)(2x+3)(1-3x)=9x\(^2\)-1
c)\(\dfrac{2x}{x+1}+\dfrac{x-1}{x}=\dfrac{2x^2+3x-1}{x^2+x}\)
a>16-x/4=2x+1/3
<=>3[16-x)=4(2x+1)
<=>48-3x=8x+8
<=>-3x-8x=8-48
<=>-5x=-40
<=>x=8
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1)\(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
