Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-\left(x+2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-x=2+4\\5x+x=-2+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
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 \(x=9\)
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) G = 9 - x2
x2\(\ge\)0
để G đạt giá trị lớn nhất
=> x2=0 x=0
=> G=9-0=9
vậy giá trị lớn nhất của G là 9
2) H = 15 - 5 . (x - 3)2
(x-3)2\(\ge\)0
=> 5.(x-3)2\(\ge\)0
để H đạt giá trị lớn nhất
=> 5.(x-3)2=0
=> H=15-0=15
vậy giá trị lớn nhất của H là 15
3) I = -y4+ 2010
I=2010 - y4
y4\(\ge\)0
để I đạt giá trị lớn nhất
=> y4=0
I= 2010-0=2010
vậy giá trị lớn nhất của I là 2010
4) J = 2015 - |x + 2014|
|x + 2014|\(\ge\)0
để J có giá trị lớn nhất
=> |x + 2014|=0
=> J=2015-0=2015
vậy giá trị lớn nhất củ J là 2015
5) K =\(\dfrac{82}{\left|x\right|+63}\)
|x|\(\ge\)0
để K đạt giá trị lớn nhất
=> |x|+63 có giá trị nhỏ nhất
=> |x|=0 x=0
|x|+63=0+63=63
=>K=82/63
6, L=\(\dfrac{\left|x\right|+20}{-29}\)
L= \(\dfrac{-20-\left|x\right|}{29}\)
|x|\(\ge\)0
để L đạt giá trị lớn nhất
=> -20-|x| đạt giá trị lớn nhất
=> |x|=0 x=0
=> -20-|x|=-20-0=-20
=>L=-20/29
Thế bạn có làm được không Võ Nguyễn Anh Thư? Trả lời thì trả lời câu hỏi ý, trả lời cái đấy để làm gì?
Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!
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
c,\(43+x=2.5^2-\left(x-57\right)\)
\(< =>43+x=50-x+57\)
\(< =>2x=50+57-43\)
\(< =>x=\frac{107-43}{2}=32\)
d,\(-3.2^2\left(x-5\right)+7\left(3-x\right)=5\)
\(< =>-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(< =>-12x+60+21-7x=5\)
\(< =>-19x=5-81=-76\)
\(< =>x=-\frac{76}{-19}=4\)
Bài 2:
a) \(A=\left|x-3\right|+10\)
Vì \(\left|x-3\right|\ge0\forall x\)\(\Rightarrow\left|x-3\right|+10\ge10\forall x\)
hay \(A\ge10\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(minA=10\Leftrightarrow x=3\)
b) \(B=-7+\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow-7+\left(x-1\right)^2\ge-7\forall x\)
hay \(B\ge-7\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=-7\Leftrightarrow x=1\)
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