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\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`
`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`
`=>x=-2014`
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)
\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> \(x+2014=0\)
\(x=0-2014\)
\(x=-2014\)
Bài 1:
Ta có:
\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)
\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)
Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:
\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)
Vậy......................
Chúc bạn học tốt!!!
Bài 2:
\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)
\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)
\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)
\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)
\(\Rightarrow-74< x< -16\)
\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)
Vậy..............................
Chúc bạn học tốt!!!
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
a) Ta có:
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=\left(x+11\right)\left(\frac{1}{15}+\frac{1}{16}\right)\)
Mà ta có:
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\ne\frac{1}{15}+\frac{1}{16}\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Ta có:
\(A=1+x+x^2+x^3+...+x^{100}\)
Đặt \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
a) \(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\) \(\Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=-2\\\dfrac{x}{2}-\dfrac{1}{3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{-5}{3}\\\dfrac{x}{2}=\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{14}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{-10}{3}\) hoặc \(x=\dfrac{14}{3}\) thì thỏa mãn đề bài.
b) \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\) \(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\) \(\Rightarrow\dfrac{x+4+2010}{2010}+\dfrac{x+3+2011}{2011}=\dfrac{x+2+2012}{2012}+\dfrac{x+1+2013}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\) \(\Rightarrow\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\) \(\Rightarrow x+2014=0\) \(\Rightarrow x=-2014\)
Vậy \(x=-2014\) thì thỏa mãn đề bài.
c) \(3^{x+2}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1+1}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}\times3+4\times3^{x+1}=7\times3^6\) \(\Rightarrow\left(3+4\right)\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}=3^6\) \(\Rightarrow x+1=6\) \(\Rightarrow x=5\)
Vậy \(x=5\) thì thỏa mãn đề bài.
a)
\(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\\ \Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=2\\\dfrac{x}{2}-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{3}+2\\\dfrac{x}{2}=\dfrac{1}{3}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{7}{3}\\\dfrac{x}{2}=\dfrac{-5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}.2\\x=\dfrac{-5}{3}.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> x + 2014 = 0
=> x = -2014
vậy x = -2014
c)\(3^{x+2}+4.3^{x+1}=7.3^6\)
\(\Rightarrow3^{x+1}.3+4.3^{x+1}=7.3^6\\ \Rightarrow3^{x+1}\left(3+4\right)=7.3^6\\ \Rightarrow3^{x+1}.7=7.3^6\\ \Rightarrow3^{x+1}=3^6\\ \Rightarrow x+1=6\\ x=6-1\\ x=5\)
vậy x = 5