giải phương trình 5-2x/3+(x-1)(x+1)/3x-1=(x+2)(1-3x)/9x-3
giải phương trình 1)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)2) \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\)3) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)4) \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)5) \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
giúp mình với ạ câu nào cũng được
giải các phương trình
a)5+(96/x^2-16)=(2x-1/x+4)-(3x-1/4-x)
b)(3x+2/3x-2)-(6/2+3x)=9x^2/9x^2-4
c)(x+1/x^2+x+1)-(x-1/x^2-x+1)=3/x(x^4+x^2+1)
a) ĐKXĐ: \(x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)
<=> 20x2 - 16x + 64 = 18x2 + 8x
<=> 20x2 - 16x + 64 - 18x2 - 8x = 0
<=> 2x2 - 24x + 64 = 0
<=> 2(x2 - 12x + 32) = 0
<=> 2(x - 8)(x - 4) = 0
<=> (x - 8)(x - 4) = 0
<=> x - 8 = 0 hoặc x - 4 = 0
<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)
=> x = 8
b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
<=> (2 + 3x)2 - 6(3x - 2) = 9x2
<=> 16 - 6x + 9x2 = 9x2
<=> 16 - 6x + 9x2 - 9x2 = 0
<=> 16 - 6x = 0
<=> -6x = 0 - 16
<=> -6x = -16
<=> x = -16/-6 = 8/3
=> x = 8/3
1.Giải phương trình:
a) 4x-8/2x^2+1 = 0
b)x^2-x-6/x-3 = 0
c)x+5/3x-6 - 1/2 = 2x-3/2x-4
d)12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x
2.Giải các phương trình:
a)5 + 96/x^2-16 = 2x-1/x+4 - 3x-1/4-x
b)3x+2/3x-2 - 6/2+3x = 9x^2/9x^2-4
c)x+1/x^2+x+1 - x-1/x^2-x+1 = 3/x(x^4+x^2+1)
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
Bài 2.
\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)
ĐK: \(x\ne\pm4\)
\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)
ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)
\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)
Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)
Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$
Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)
Giải phương trình
1, \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
2, \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)3, \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)4, \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)5, \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
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 : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
Giải Phương Trình
a, (2x+3)^2 - 3(x-4) (x+4) = (x-2)^2 +1
b, (3x-2) (9x^2 + 6x +4) - (3x-1) ( 9x^2 - 3x +1) = x+4
c, x(x-1) - (x-3) (x+4) = 5x
d, (2x+1) (2x-1) = 4x(x-7) - 3x
(3x-1)(2x+7) + (x-1)(5-6x) =16
\(\left(x^4+3x^3+9x-9\right):\left(x^2+3\right)=-3\)
(x+1)(x^2-x+1) - (x-3) (x+3) =8
giải phương trình
Giải phương trình:
A) 1-x/x+1 +3 = 2x+3/x+1
B) (x+2)^2/2x-3 -1 = x^2-10/2x-3
C) 5x-2/2-2x + 2x-1/2 = 1 + x^2+x-3/x-1
D) 5-2x/3 - (x-1)(x+1)/1-3x = (x+2)(1-3x)/9x-3
E) x-3/x-2 + x-2/x-4 = -1
F) 1 + x/3-x = 5/(x+2)(3-x) + 2/x+2
G) x+1/x-1 - x-1/x+1 = 3x( 1 - x-1/x+1 )
H) 1-6x/x-2 + 9x-4/x+2 = x(3x-2)+1/x^2-4
I) 3x-1/x-1 - 2x+5/x+3 + 4/x^2+2x-3 = 1
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
Trả lời:
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)\(\left(đkxđ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow4+2x=2x+3\)
\(\Leftrightarrow2x-2x=3-4\)
\(\Leftrightarrow0x=-1\)(không thỏa mãn)
Vậy \(S=\varnothing\)
b, \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\)\(\left(đkxđ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{\left(x+2\right)^2-\left(2x-3\right)}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow x^2+2x+7=x^2-10\)
\(\Leftrightarrow x^2+2x-x^2=-10-7\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(tm)
Vậy \(S=\left\{\frac{-17}{2}\right\}\)
c, \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)\(\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\frac{2-5x}{2x-2}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)
\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)
\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)}+\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)
\(\Rightarrow2-5x+2x^2-3x+1=2x-2+2x^2+2x-6\)
\(\Leftrightarrow2x^2-8x+3=2x^2+4x-8\)
\(\Leftrightarrow2x^2-8x-2x^2-4x=-8-3\)
\(\Leftrightarrow-12x=-13\)
\(\Leftrightarrow x=\frac{13}{12}\)(tm)
Vậy \(S=\left\{\frac{13}{12}\right\}\)
1.Giải phương trình:
a) 4x-8/2x^2+1 = 0
b)x^2-x-6/x-3 = 0
c)x+5/3x-6 - 1/2 = 2x-3/2x-4
d)12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x
2.Giải các phương trình:
a)5 + 96/x^2-16 = 2x-1/x+4 - 3x-1/4-x
b)3x+2/3x-2 - 6/2+3x = 9x^2/9x^2-4
c)x+1/x^2+x+1 - x-1/x^2-x+1 = 3/x(x^4+x^2+1)
(mong mn giúp mk, mk đang thật sự gấp, cảm ơn mọi người rất nhiều)
\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình trên là \(-2\)
\(c.\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\left(x\ne2\right)\\ \Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\\\Leftrightarrow \frac{2\left(x+5\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\\\Leftrightarrow 2\left(x+5\right)-3\left(x-2\right)=3\left(2x-3\right)\\\Leftrightarrow 2x+10-3x+6=6x-9\\\Leftrightarrow 2x-3x-6x=-10-6-9\\\Leftrightarrow -7x=-25\\\Leftrightarrow x=\frac{25}{7}\left(tm\right)\)
Vậy nghiệm của phương trình trên là \(\frac{25}{7}\)