1 phần 2 ( x + 1 phần 4 ) - 1 phần 3 =4 phần 3
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\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 4:so sánh
5/2 lớn hơn 3/7
4/3 lớn hơn,3/2 lớn hơn
bài 6:rút gọn các phân số sau:
3/9=1/3 9/12=3/4 8/18=4/9 60/36=10/6 17/34=1/2 17/51=1/3 35/100=7/20 25/100=1/4 8/1000=1/125 24/30=4/5 18/54=1/3 72/42=12/7
đay nhé mk chưa làm hết đc bn viết liền quá mk nhìn khó mà mk hỏi bài 7 là nhân hay cộng vậy?
![](https://rs.olm.vn/images/avt/0.png?1311)
3 x \(\frac{4}{11}\) = \(\frac{3}{1}\) x \(\frac{4}{11}\) = \(\frac{12}{11}\)
1 : \(\frac{5}{4}\) = \(\frac{1\times4}{5}\) = \(\frac{4}{5}\)
\(\frac{4}{5}\) x \(\frac{6}{7}\) + \(\frac{2}{3}\) = \(\frac{24}{35}\) + \(\frac{2}{3}\) = \(\frac{142}{105}\)
\(\frac{3}{4}\) + \(\frac{1}{2}\) : \(\frac{5}{6}\) = \(\frac{3}{4}\) + \(\frac{1}{2}\) x \(\frac{6}{5}\) = \(\frac{3}{4}\)+ \(\frac{6}{10}\) \(\frac{27}{20}\)
\(\frac{3}{4}\)+ \(\frac{1}{2}\) = \(\frac{3}{4}\) + \(\frac{2}{4}\)= \(\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm :
1)\(\frac{3}{2}+\frac{5}{2}-\frac{4}{2}=\frac{3+5-4}{2}=\frac{4}{2}=2\)
2)\(\frac{2}{5}\times\frac{1}{2}\div\frac{1}{3}=\frac{2\times1\times3}{5\times2\times1}=\frac{6}{10}=\frac{3}{5}\)
3)\(\frac{2}{9}\div\frac{9}{2}\times\frac{1}{2}=\frac{2\times2\times1}{9\times9\times2}=\frac{4}{162}=\frac{2}{81}\)
4)\(\left(\frac{4}{11}+\frac{3}{11}\right)\times\frac{2}{6}=\frac{4+3}{11}\times\frac{2}{6}=\frac{7}{11}\times\frac{2}{6}=\frac{7\times2}{11\times6}=\frac{14}{66}=\frac{7}{33}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
Ta có : \(\frac{1}{2}.\left(x+\frac{1}{4}\right)-\frac{1}{3}=\frac{4}{3}\)
\(\Rightarrow\) \(\frac{1}{2}.\left(x+\frac{1}{4}\right)=\frac{4}{3}+\frac{1}{3}\)
\(\Rightarrow\) \(\frac{1}{2}.\left(x+\frac{1}{4}\right)=\frac{5}{3}\)
\(\Rightarrow\) \(x+\frac{1}{4}=\frac{5}{3}:\frac{1}{2}\)
\(\Rightarrow\) \(x+\frac{1}{4}=\frac{10}{3}\)
\(\Rightarrow\) \(x=\frac{10}{3}-\frac{1}{4}\)
\(\Rightarrow\) \(x=\frac{37}{12}\)
Vậy \(x=\frac{37}{12}\)
Tk và Kb vs mk nhoa ✿
=> \(\frac{1}{2}\cdot\left(x+\frac{1}{4}\right)-\frac{1}{3}=\frac{4}{3}\)
=> \(\frac{1}{2}\cdot\left(x+\frac{1}{4}\right)=\frac{4}{3}+\frac{1}{3}\)
=> \(\frac{1}{2}\cdot\left(x+\frac{1}{4}\right)=\frac{5}{3}\)
=> \(x+\frac{1}{4}=\frac{5}{3}\div\frac{1}{2}\)
=> \(x+\frac{1}{4}=\frac{10}{3}\)
=> \(x=\frac{10}{3}-\frac{1}{4}\)
=> \(x=\frac{37}{12}\)
Vậy \(x=\frac{37}{12}\)