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\(\left[1,3\right]=1\)
\(\left[-2,4\right]=-3\)
\(\left[-3\right]=-3\)
\(\left[-4\right]=-4\)
\(\left[1\right]=1\)
\(\left[-6\right]=-6\)
\(\left[25\right]=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.
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Bài 1 :Bỏ dấu ngoặc
2007-(7-3+4)
= 2007 -7+3-4
= 1999
6+[(-5) + 4 - 1 ]
= 6-5+4-1
=4
5-[(-6+8-2]
= 5+6-8+2
=5
-10+(7-3+1)
= -10 +7-3+1
= -5
Bài 3 Tìm x
\(\dfrac{1}{3} = \dfrac{x}{6}\)
\(<=> x= \dfrac{1.6}{3}\)
\(<=> x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}-\dfrac{2}{5}\right)+\left(-1\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(-x+\dfrac{1}{10}\right)=\dfrac{3}{10}\)
\(\left(\dfrac{5}{10}-\dfrac{4}{10}\right)+\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+x-\dfrac{1}{10}=\dfrac{3}{10}\)
\(\dfrac{1}{10}+\left(\dfrac{-4}{6}-\dfrac{5}{6}\right)+x=\dfrac{3}{10}+\dfrac{1}{10}\)
\(\dfrac{1}{10}+\dfrac{-3}{2}+x=\dfrac{4}{10}\)
\(\dfrac{-3}{2}+x=\dfrac{4}{10}-\dfrac{1}{10}\)
\(\dfrac{-3}{2}+x=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}-\dfrac{-3}{2}\)
\(x=\dfrac{3}{10}-\dfrac{-15}{10}\)
\(x=\dfrac{18}{10}=\dfrac{9}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: =>x-1/5=2/15
=>x=2/15+3/15=5/15=1/3
b: =>x+7/12=-5/6-2/6=-7/6
=>x=-14/12-7/12=-21/12=-7/4
c: =>x+2/3=-10/3
=>x=-4
d: =>1/4:x=-11/4
=>x=-1/4:11/4=-1/11
e: =>8:x=1,6
=>x=5