Giải phương trình:
\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
a)
PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)
\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)
b)
PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)
\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)
Vậy pt vô nghiệm.
Giải phương trình:
\(\dfrac{3x-1}{x-1}\) - \(\dfrac{2x+5}{x+3}\) + \(\dfrac{4}{x^2+2x-3}\) = 1
`(3x-1)/(x-1)-(2x+5)/(x+3)+4/(x^2+2x-3)=1(x ne 1,-3)`
`<=>((3x-1)(x+3))/(x^2+2x-3)-((2x+5)(x-1))/(x^2+2x-3)+4/(x^2+2x-3)=(x^2+2x-3)/(x^2+2x-3)`
`<=>(3x-1)(x+3)-(2x+5)(x-1)+4=x^2+2x-3`
`<=>3x^2+8x-3-2x^2-3x+5+4=x^2+2x-3`
`<=>x^2+5x+6=x^2+2x-3`
`<=>3x=-9`
`<=>x=-3(loại)`
Vậy `S={cancel0}`
ĐKXĐ: \(x\notin\left\{1;-3\right\}\)
Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}+\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3x^2+8x-3-\left(2x^2+3x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{3x^2+8x+1-2x^2-3x+5}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(x^2+5x+6-x^2-2x+3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3x=-9\)
hay x=-3(Không nhận)
Vậy: \(S=\varnothing\)
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 ẩn x sau:
a) \(\dfrac{1}{3x}\)+\(\dfrac{1}{2x}\)=\(\dfrac{1}{4}\)
b) \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
c)\(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
d) \(\dfrac{2a}{x+a}=1\)
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{3x}+\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)
Suy ra: \(3x=10\)
\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)
b) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)
Suy ra: \(3x-4x=8\)
\(\Leftrightarrow-x=8\)
hay x=-8(thỏa ĐK)
Vậy: S={-8}
c)ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)
Suy ra: 2x+3x=10
\(\Leftrightarrow5x=10\)
hay x=2(thỏa ĐK)
Vậy: S={2}
d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)
\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)
\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)
\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)
\(\Leftrightarrow\) x = a (TM)
Vậy S = {a}
Chúc bn học tốt!
Giải các phương trình
1,\(3x-1=0\\\) 4, \(\dfrac{x}{3}-\dfrac{x}{5}=4\)
2, \(2-x=3x+1\) 5, \(\dfrac{x-1}{4}+\dfrac{2x+1}{6}=\dfrac{3}{2}\)
3, \(2\left(x-2\right)-1=5x\)
1,\(3x-1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)
2,\(2-x=3x+1\Leftrightarrow2-1=3x+x\rightarrow1=4x\Rightarrow x=-\dfrac{1}{4}\)
3,\(2\left(x-2\right)-1=5x\Leftrightarrow2x-4-1=5x\Leftrightarrow2x-5x=4+1\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)
4,\(\dfrac{x}{3}-\dfrac{x}{5}=4\Leftrightarrow\dfrac{5x}{15}-\dfrac{3x}{15}=\dfrac{60}{15}\Rightarrow5x-3x=60\Rightarrow2x=60\Rightarrow x=\dfrac{60}{2}=30\)
5,\(\dfrac{x-1}{4}+\dfrac{2x+1}{6}=\dfrac{3}{2}\Leftrightarrow\dfrac{3\left(x-1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{18}{12}\)
\(3\left(x-1\right)+2\left(2x+1\right)=18\Leftrightarrow3x-3+4x+2=18\Leftrightarrow3x+4x=3-2+18\Rightarrow7x=19\Rightarrow x=\dfrac{19}{2}\)
Giải phương trình ;
\(\dfrac{2x+5}{x+3}\)+ 1 = \(\dfrac{4}{x^3+2x-3}\)+ \(\dfrac{9-3x}{1-x}\)
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)
2) \(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
3) \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10}\)
4) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
5) \(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
mng giúp mk bài này nha. Cảm ơn bạn nhiều
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
Giải các phương trình sau:
\(h.\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)
\(\Leftrightarrow15x+9x-1=14-7x\)
\(\Leftrightarrow31x=15\)
\(\Leftrightarrow x=\dfrac{15}{31}\)
giải các phương trình sau:
a)2x(x-2)+5(x-2)=0
b)\(\dfrac{3x-4}{2}-\dfrac{4x+1}{3}\)
c)\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)
a: =>(x-2)(2x+5)=0
=>x-2=0 hoặc 2x+5=0
=>x=2 hoặc x=-5/2
c: \(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)
=>\(\dfrac{2x^2+2x-x^2+x}{x^2-1}=1\)
=>x^2+3x=x^2-1
=>3x=-1
=>x=-1/3
giải các phương trình sau:
a)2x(x-2)+5(x-2)=0
b)\(\dfrac{3x-4}{2}\)-\(\dfrac{4x+1}{3}\)
c)\(\dfrac{2x}{x-1}\)-\(\dfrac{x}{x+1}=1\)
\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)
\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\) ( ĐKXĐ: \(x\ne-1;x\ne1\) )
\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\) ( nhận )
Vậy phương trình có tập nghiệm S = \(\left\{-\dfrac{1}{3}\right\}\)