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\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+6-4}=\frac{2x-2+3y-6-z+3}{4+6-4}\)
\(=\frac{\left(2x+3y-z\right)+\left(-2+6+3\right)}{6}=\frac{50+\left(-5\right)}{6}=\frac{45}{6}=7,5\)
\(\frac{x-1}{2}=7,5\Rightarrow x-1=15\Rightarrow x=16\)
\(\frac{y-2}{3}=7,5\Rightarrow y-2=24,5\Rightarrow y=20,5\)
\(\frac{z-3}{4}=7,5\Rightarrow z-3=30\Rightarrow z=33\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(=>\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(=>\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(=>\left(\frac{x+2016}{2015}+\frac{x+2016}{2014}\right)-\left(\frac{x+2016}{2013}+\frac{x+2016}{2012}\right)=0\)
\(=>\left(x+2016\right).\left[\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)\right]=0\)
\(=>\orbr{\begin{cases}x+2016=0\\\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)=0\end{cases}}\)
Do 1/2015 + 1/2014 < 1/2013 + 1/2012
=> (1/2015 + 1/2014) - (1/2013 + 1/2012) khác 0
=> x - 2016 = 0
=> x = 2016
Vậy x = 2016
Ủng hộ mk nha ^_-
Ta có: \(\frac{x}{4}=\frac{27}{x}\Leftrightarrow x^2=108\Leftrightarrow x=\sqrt{108}\)
\(\left(3x-1\right)^3=\frac{-8}{27}\)
Mà \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\Rightarrow3x=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)
Vậy x = 1/9
+) \(\frac{x}{4}=\frac{27}{x}\)
\(\Rightarrow x\times x=27\times4\)
\(\Rightarrow x^2=108\)
\(\Rightarrow x=\sqrt{108}\)
Vậy \(x=\sqrt{108}\)
+) \(\left(3x-1\right)^3=\frac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{-2}{3}\)
\(\Rightarrow3x=\frac{-2}{3}+1\)
\(\Rightarrow3x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:3\)
\(\Rightarrow x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
_Chúc bạn học tốt_
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(\Rightarrow x=x\)hoặc \(x=-x\)
---Nếu x = x
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{2}{3}\right|+\left|x+\frac{3}{4}\right|=4x\)
\(x+\frac{1}{2}+x+\frac{2}{3}+x+\frac{3}{4}=4x\)
\(3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x-3x\)
\(\Rightarrow x=\frac{23}{12}\)
---Nếu x = -x
\(\Rightarrow\left|-x+\frac{1}{2}\right|+\left|-x+\frac{2}{3}\right|+\left|-x+\frac{3}{4}\right|=4x\)
\(-x+\frac{1}{2}+\left(-x\right)+\frac{2}{3}+\left(-x\right)+\frac{3}{4}=4x\)
\(-3x+\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=4x\)
\(\left(\frac{6}{12}+\frac{8}{12}+\frac{9}{12}\right)=4x+3x\)
\(\Rightarrow7x=\frac{23}{12}\)
\(\Rightarrow x=\frac{23}{12}:7\)
\(\Rightarrow x=\frac{23}{12}.\frac{1}{7}\)
\(\Rightarrow x=\frac{23}{84}\)
\(Vậyx=\frac{23}{12};x=\frac{23}{84}\)