\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
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Những câu hỏi liên quan
18 tháng 9 2021
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
GH
11 tháng 2 2018
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
H
16 tháng 10 2023
`3/5 : x =1/3 +1/2`
` 3/5 : x= 2/6 +3/6`
` 3/5 : x= 5/6`
` x= 3/5 : 5/6`
` x= 3/5 xx 6/5`
` x= 18/25`
__
`x: 7/15 =2`
` x= 2xx 7/15`
` x= 14/15`
__
`x-3/2=11/4-5/4`
`x-3/2= 6/4`
`x= 3/2 +3/2`
`x= 6/2`
`x=3`
__
`x+5/4 = 3/2+7/12`
`x+5/4 = 18/12+7/12`
`x+5/4 = 25/12`
`x= 25/12-5/4`
`x= 25/12- 15/12`
`x= 10/12`
`x= 5/6`
11 tháng 8 2021
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
24 tháng 8 2021
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
24 tháng 8 2021
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
S
2
MH
10 tháng 2 2022
\(\dfrac{1}{x+2}+\dfrac{6x+12}{x^3+8}-\dfrac{7}{x^2-2x+4}=0\) \(\left(đk:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{x^2-2x+4+6x+12-7\left(x+2\right)}{x^3+8}=0\)
\(\Leftrightarrow\dfrac{x^2-3x+2}{x^3+8}=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(TM)
Vậy ...
10 tháng 2 2022
dk : x khac -2
\(\Rightarrow x^2-2x+4+6x+12-7\left(x+2\right)=0\)
\(\Leftrightarrow x^2+4x+16-7x-14=0\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2\)
26 tháng 6 2017
a, \(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}\right)=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+7\right)}{\left(x-2\right)\left(x+2\right)}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{x^2+3x+2-x^2-5x+14}{x^2-4}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{14-2x}{x^2-4}\right)=0\)
\(\Leftrightarrow12=14-2x\)
\(\Leftrightarrow x=1\)
Vậy x = 1
DB
1
29 tháng 6 2021
`h)x/2-1/x=1/12(x ne 0)`
`<=>6x^2-12=x`
`<=>6x^2-x-12=0`
`<=>6x^2-9x+8x-12=0`
`<=>3x(2x-3)+4(2x-3)=0`
`<=>(2x-3)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\)
`i)x^2-7/6x+1/3=0`
`<=>6x^2-7x+2=0`
`<=>6x^2-3x-4x+2=0`
`<=>3x(2x-1)-2(2x-1)=0`
`<=>(2x-1)(3x-2)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\)
Câu cuối không có dấu "=" nên không tìm được x :v
29 tháng 6 2021
- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v
ĐK: `x \ne \pm 2`
`12/(x^2-4)-(x+1)/(x-2)+(x+7)/(x+2)=0`
`<=>12-(x+1)(x+2)+(x+7)(x-2)=0`
`<=> 2x-4=0`
`<=> x=2 (L)`
Vậy `S={∅}`.
\(x\ne\pm2\)
\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=> \(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow12-x^2-x-2x-2+x^2+7x-2x-14=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow x=2\) (ktm)
Vậy pt vô nghiệm.