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1. |x| - x = 0
<=> |x| = x
<=> \(\left[{}\begin{matrix}x=x\\x=-x\end{matrix}\right.\) (thỏa mãn)
@Phan Đức Gia Linh
2. |x| + x = 0
<=> |x| = -x
Do |x| \(\ge\) 0, mà -x < 0 => không tồn tại x thỏa mãn
@Phan Đức Gia Linh
1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)
2) thiếu quế phải nha
3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)
\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)
\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)
4) ta có : \(x+y=20\Leftrightarrow y=20-x\)
\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)
\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)
\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)
\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)
\(\dfrac{23+x}{40-x}=\dfrac{-3}{4}\Leftrightarrow4\left(23+x\right)=-3\left(40-x\right)\)
\(\Leftrightarrow92+4x=-120+3x\Leftrightarrow4x-3x=-120-92\)
\(\Leftrightarrow x=-212\) vậy \(x=-212\)
\(\left|x-2\right|+\left|2y-5\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|2y-5\right|\ge0\forall y\end{matrix}\right.\)
\(\left|x-2\right|+\left|2y-5\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2\right|=0\Rightarrow x=2\\\left|2y-5\right|=0\Rightarrow2y=5\Rightarrow y=\dfrac{5}{2}\end{matrix}\right.\)
\(\left|3y-2\right|+\left|xy-6\right|=0\)
\(\left\{{}\begin{matrix} \left|3y-2\right|\ge0\forall y\\\left|xy-6\right|\ge0\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left|3y-2\right|+\left|xy-6\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|3y-2\right|=0\Rightarrow3y=2\Rightarrow y=\dfrac{3}{2}\\\left|xy-6\right|=0\Rightarrow\dfrac{3}{2}x=6\Rightarrow x=4\end{matrix}\right.\)
\(\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|2y-\dfrac{1}{3}\right|\ge0\forall y\\ \left|4z-5\right|\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\\\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\Rightarrow x=\dfrac{1}{2}\\\left|2y-\dfrac{1}{3}\right|=0\Rightarrow2y=\dfrac{1}{3}\Rightarrow y=\dfrac{1}{6}\\\left|4z-5\right|=0\Rightarrow4z=5\Rightarrow z=\dfrac{5}{4}\end{matrix}\right.\)
1. x3 - \(\dfrac{4}{25}\)x = 0
<=> x(x2 - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh
1 ) \(x^3-\dfrac{4}{25}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)
Vậy .............
2 ) \(3^{4x+4}=9^{x+2}\)
\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)
\(\Leftrightarrow4x+4=2x+4\)
\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)
3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )
4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)
\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)
Vậy ......
5) \(2+4+6+...+2x=210\)
\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)
\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)
\(\Leftrightarrow1+2+3+...+x=105\)
\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)
\(\Leftrightarrow x\left(x+1\right)=210\)
Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)
\(\Leftrightarrow x=14\)
Vậy ......
6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x=17.\)
Vậy ...........
\(\)
1: \(x^2\left(2-x\right)\le0\)
\(\Leftrightarrow2-x\le0\)
hay x>=2
2: \(\left(x-7\right)\left(x+3\right)< 0\)
=>x+3>0 và x-7<0
=>-3<x<7
3: \(\left(x+4\right)\left(x-3\right)>0\)
=>x-3>0 hoặc x+4<0
=>x>3 hoặc x<-4
1: \(A=x^2+y^2+2014\ge2014\)
Dấu '=' xảy ra khi x=y=0
2: \(B=\left(x+30\right)^2+\left(y-4\right)^2+17\ge17\)
Dấu '=' xảy ra khi x=-30 và y=4
3: \(C=\left(y-9\right)^2+\left|x-3\right|-1\ge-1\)
Dấu '=' xảy ra khi x=3 và y=9
\(1\text{)} \left(x-3\right)-9.\left(2x-7\right)=16:2\\ x-3-18x+63=8\\ -17x=-52\\ x=\dfrac{52}{17}\)
\(2\text{)} x-\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\\ -\dfrac{1}{6}x=\dfrac{5}{12}\\ x=-2,5\)
\(3\text{)} -2.\left(x-1\right)+5.\left(x-2\right)=x-14\\ -2x+2+5x-10=x-14\\ 2x=-6\\ x=-3\)
\(4\text{)} x.\left(x+1\right)-x-1=0\\ \left(x+1\right)\left(x-1\right)=0\\ x^2-1=0\\ x^2=1\\ x=\pm1\)
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
1) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-\left(x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-x=2+4\\5x+x=-2+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{3}{2}\)
2) \(\left|x+15\right|=\left|3x-4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-\left(3x-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-3x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-4-15\\x+3x=4-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=-19\\4x=-11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
Vậy \(x=-\dfrac{11}{4}\) hoặc \(x=\dfrac{19}{2}\)
3) \(\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|-\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|=\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\left(\dfrac{5}{8}x+\dfrac{3}{5}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{8}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy \(x=\dfrac{116}{75}\) hoặc \(x=\dfrac{164}{25}\)
4) \(\left|2x-6\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|2x-6\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-\left(x+3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3+6\\2x+x=-3+6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\3x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=9\)
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