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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 \(x=-212\)
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. 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 ...........
\(\)
\(2^x.4^2-2^{x+1}=2^6-2^3\)
\(2^x.2^4-2^x.2=2^2.\left(2^4-2\right)\)
\(2^x.\left(2^4-2\right)=2^2.\left(2^4-2\right)\)
\(2^x.14=2^2.14\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3^{50}=\left(3^5\right)^{10}=243^{10}\)
\(5^{30}=\left(5^3\right)^{10}=125^{10}\)
Ta có: \(125< 243\)
\(\Rightarrow125^{10}< 243^{10}\)
\(\Rightarrow3^{50}< 5^{30}\)
\(\left|x-2\right|+\left|2y-5\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|2y-5\right|\ge0\forall y\end{matrix}\right.\)
\(\left|x-2\right|+\left|2y-5\right|\ge0\)
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:
\(\left\{{}\begin{matrix}\left|3y-2\right|=0\Rightarrow3y=2\Rightarrow y=\dfrac{3}{2}\\\left|xy-6\right|=0\Rightarrow\dfrac{3}{2}x=6\Rightarrow x=4\end{matrix}\right.\)
\(\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|2y-\dfrac{1}{3}\right|\ge0\forall y\\ \left|4z-5\right|\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\\\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\Rightarrow x=\dfrac{1}{2}\\\left|2y-\dfrac{1}{3}\right|=0\Rightarrow2y=\dfrac{1}{3}\Rightarrow y=\dfrac{1}{6}\\\left|4z-5\right|=0\Rightarrow4z=5\Rightarrow z=\dfrac{5}{4}\end{matrix}\right.\)
a/ \(3^x-3^{x+3}=-234\)
\(\Rightarrow3^x\left(1-3^3\right)=-234\)
\(\Rightarrow3^x=\dfrac{-234}{-26}=9=3^2\)
\(\Rightarrow x=2\)
b/ đề có sai k bn>?
a, \(3^x-3^{x+3}=-234\)
\(3^x\left(1-3^3\right)=-234\)
\(3^x.\left(-26\right)=-234\)
\(3^x=-234:\left(-26\right)\)
\(3^x=9\)
\(3^x=3^2\Rightarrow x=2\)
b, Mk sửa lại đề:
\(3.3^x+5.3^{x+1}=162\) \(\)
\(3^{x+1}+5.3^{x+1}=162\)
\(3^{x+1}\left(1+5\right)=162\)
\(3^{x+1}.6=162\)
\(3^{x+1}=162:6\)
\(3^{x+1}=27\)
\(3^{x+1}=3^3\)
\(\Rightarrow x+1=3\)
\(x=3-1\)
\(\Rightarrow x=2\)