\(\dfrac{1}{(x-3)(x-4)}\)\(+\dfrac{3}{(x-4)(x-7)}+\dfrac{5}{(x-7)(x-12)}=\dfrac{1}{4}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`
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`x: 7/15 =2`
` x= 2xx 7/15`
` x= 14/15`
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`x-3/2=11/4-5/4`
`x-3/2= 6/4`
`x= 3/2 +3/2`
`x= 6/2`
`x=3`
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`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`
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=-\dfrac{11}{3}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
60x [7/12+4/15]
60x153/180
=9180/180
b 1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9=40320/4032
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,sửa đề : đk x khác -2; 2
\(x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)
b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)
c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)
\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)
a. ko hỉu đề lắm :v
b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)
\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)
\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)
\(\Leftrightarrow3x-12+5+5x-105=0\)
\(\Leftrightarrow8x-112=0\)
\(\Leftrightarrow8x=112\)
\(\Leftrightarrow x=14\)
c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)
\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)
\(\Leftrightarrow4x^2+16x-64=0\)
Nghiệm xấu lắm bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107`
a)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)
Vậy, \(x=-\dfrac{1}{5}\)
b)
\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)
Vậy, \(x=\dfrac{147}{20}\)
c)
\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)
Vậy, \(x=-\dfrac{1}{3}.\)
\(#Emyeu1aithatroi...\)
(2/5 + 3/4 . x)= 11/12 -2/3
(2/5 +3/4 . x)= 1/4
3/4 . x = 1/4 - 2/5
3/4 . x = -3/20
x = -3/20 : 3/4
x = -1/5
Vậy .....
ĐKXĐ:\(x\ne3,4,7,12\)
\(\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{3}{\left(x-4\right)\left(x-7\right)}+\dfrac{5}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{x-12-x+3}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{-9}{x^2-15x+36}=\dfrac{1}{4}\\ \Leftrightarrow x^2-15x+36=-36\\ \Leftrightarrow x^2-15x+72=0\)
\(\Leftrightarrow x^2-2.\dfrac{15}{2}x+\dfrac{225}{4}+\dfrac{63}{4}=0\\ \Leftrightarrow\left(x-\dfrac{15}{2}\right)^2+\dfrac{63}{4}=0\left(vô.lí\right)\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-12}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{x-12-x+3}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-3\right)\left(x-12\right)=-36\)
=>x(x-15)=0
=>x=0 hoặc x=15