a.2(x-3)=4-2x
b.3x(x-2)=3(x-2)
c.(2x-1)^2=25
d.\(\frac{x+1}{2x-2}-\frac{x-1}{2x+2}+\frac{2}{1-x^2}=0\)
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Bµi 5: Gi¶i PT sau.
\(a,\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
b,\(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
\(c,\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
d) (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
e) x4 + 2x3 + 4x2 + 2x + 1 = 0
\(f,\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)
a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)
ĐK: x≠1
<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)
<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)
<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)
<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)
<=> 10x-9=2(1-x)
<=>10x-9=2-2x
<=> 10x+2x= 2+9
<=> 12x=11
<=> x= \(\frac{11}{12}\left(tm\right)\)
b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)
ĐK: x≠2, x≠-2
<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)
<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x2-2x+1)=0
<=> -(6x2-x+12x-2)+9x2+4x-18x-8-3x2+2x-1 = 0
<=> -6x2-11x+2+9x2+4x-18x-8-3x2+2x-1=0
<=> -23x-7=0
<=> -23x=7
<=> x= \(\frac{-7}{23}\left(tm\right)\)
tham khảo câu d trong
https://hoc24.vn/hoi-dap/question/919967.html
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
\(1.\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}
\)
2.\(\frac{2x^4}{\left(x+1\right)^2}-\frac{5x^2}{x+1}+2=0\)
3.\(\left(x+\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+8=0\)
4.\(\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0\)
5.\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
bài 1
a \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{2x^2}{7x-10-x^2}\)=0
b\(\frac{2x+5}{x+3}-\frac{2x+2}{\left(x-1\right)\left(x+3\right)}=\frac{3x-1}{x-1}\)
c (2x-1)\(^2\)- (2x+1)\(^2\)=4(x-3)
a) Ta có: \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{2x^2}{7x-10-x^2}=0\)
\(\Leftrightarrow\frac{3x}{x-2}-\frac{x}{x-5}-\frac{2x^2}{x^2-7x+10}=0\)
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}-\frac{2x^2}{\left(x-5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow3x^2-15x-x^2+2x-2x^2=0\)
\(\Leftrightarrow-13x=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
c) Ta có: \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow-8x-4x+12=0\)
\(\Leftrightarrow-12x+12=0\)
\(\Leftrightarrow x=\frac{-12}{-12}=1\)
Vậy: x=1
\(\frac{x^2-x}{x^2-x+1}-\frac{x^2-x+2}{x^2-x-2}=1.\)
\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
\(\frac{1}{x^2-2x+2}+\frac{1}{x^2-2x+3}=\frac{9}{2\left(x^2-2x+4\right)}\)
\(\frac{1}{x^2-2x+3}+\frac{1}{x^2-2x+2}=\frac{6}{x^2-2x+4}\)
mai mik kiểm tra rùi giúp mik vs pls
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Giải các phương trình.
a) \(\frac{2.\left(1-3x\right)}{5}-\frac{2+3x}{10}=7-\frac{3.\left(2x+1\right)}{4}\)b) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
c) 3x-5=7
d) \(\frac{5}{x+3}=\frac{3}{x-1}\)
e) -2x+14=0
f) 2x.(x-3)+5.(x-3)=0
g) (x2-4)-(x-2).(3-2x)=0
h) 2x3+6x2=x2+3x
giải ác phương trình sau:
1)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=0\)
2)\(\frac{x}{x-1}-\frac{5x-3}{x^2-1}=0\)
3)\(\frac{1}{x-3}-\frac{4}{x+3}=\frac{3x}{9-x^2}\)
4)\(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
5)\(\frac{-3}{2x}-\frac{x+1}{x+2}=\frac{-3}{x\left(x+2\right)}\)
6)\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
1, Đk x≠2;-2
\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=0\\ =>\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{\left(x+2\right)^2}{2\left(x^2-4\right)}-\frac{8x}{2\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x-2}{2\left(x+2\right)}=0\\ =>x-2=0\\ =>x=2\left(loại\right)\)
1) tính
a) \(\frac{4}{x+2}+\frac{3}{2-x}+\frac{12}{x^2-4}\)
b) \(x+\frac{x-1}{2}+\frac{x-2}{3}\)
c) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)
d) x - 2 - \(\frac{x^2-10}{x+2}\)
e) \(\frac{1}{2x-2y}-\frac{1}{2x+2y}+\frac{y}{y^2-x^2}\)
f) \(\frac{1}{a+1}-\frac{3}{a^3+1}+\frac{3}{a^2-a+1}\)
g) \(\frac{4-2x+x^2}{x+2}-2-x\)
h)\(\frac{1}{x^3-x}-\frac{1}{x^2-x}+\frac{2}{x^2-1}\)
j) \(\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2-x-3}\)
a: \(=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
b: \(=\dfrac{6x+3\left(x-1\right)+2\left(x-2\right)}{6}=\dfrac{6x+3x-3+2x-4}{6}=\dfrac{11x-7}{6}\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)