\(\frac{x-2}{3}\)-\(\frac{x+2}{4}\)
ai giúp mình với
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\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)
=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396
=>4x+10=-396
4x=-406
x=-406:4=-101,5
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Theo đề bài cho => \(4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}=4-\frac{5}{x}\)
=> \(x=6-\frac{7}{8-\frac{9}{10}}\)
Vậy x=...
Nếu thế mk đúng thì ủng hộ nha
Ta có:\(2-\frac{3}{4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{7}{\frac{71}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{70}{71}}}\)
\(=2-\frac{3}{4-\frac{5}{\frac{356}{71}}}\)
=>x=\(\frac{356}{71}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
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\(\frac{x+4}{2009}+\frac{x+3}{2010}=\frac{x+2}{2011}+\frac{x+1}{2012}\)\(\Leftrightarrow\)\(\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+1}{2012}+1\right)\)
\(=\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\)
Biểu thức trên chi thỏa mãn khi x+2013=0
\(\Rightarrow x=-2013\)
mk nghĩ là -2013 vì nếu thay x=-2013 vào thì các phân số sẽ bằng -1.
nếu cộng lại thì đc -2
k nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đk : \(x\ne4,-4\)
A= \(\frac{8+x-4}{\left(x+4\right)\left(x-4\right)}:\frac{2\left(x-4\right)-x^2}{2x\left(x+4\right)}\)
A = \(\frac{x+4}{\left(x-4\right)\left(x+4\right)}.\frac{2x\left(x+4\right)}{x^2+2x-8}\)
A=\(\frac{1}{x-4}.\frac{2x\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{2x}{\left(x-4\right)\left(x-2\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)
\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)