tim x biet -\left|2x+3\right|\ge 0
Tim m de pt \(x^2-2x-2\left|x-m\right|+1=0\) co 3 nghiem phan biet
Tim x,biet;
a/ \(\left(x-1\right)^2=0\)
b/ \(\left(x-2\right)^2-1=0\)
c/\(\left(2x-1\right)^3=-8\)
d/ \(\left(x+2\right)^2+1=0\)
a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)
vậy phương trình vô nghiệm
a) (x-1)2 = 0
<=> x-1 = 0
<=> x = 1
b) (x-2)2 - 1 = 0
<=> (x-2)2 = 1
<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) (2x-1)3 = -8
<=> (2x-1)3 = -23
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(-\dfrac{1}{2}\)
d) (x+2)2 + 1 = 0
<=> (x+2)2 = -1
<=> x+2 = -1
<=> x = -3
a, \(\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2=0^2\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ......
b, \(\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy .....
c, \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy ...
d, \(\left(x+2\right)^2+1=0\)
\(\Leftrightarrow\left(x+2\right)^2=-1\)
\(\Leftrightarrow\) ko tìm dc giá trị của x thỏa mãn (do \(\left(x+2\right)^2\ge0\))
Giải các bất phương trình sau:
a) \(0,{1^{2 - x}} > 0,{1^{4 + 2x}};\)
b) \({2.5^{2x + 1}} \le 3;\)
c) \({\log _3}\left( {x + 7} \right) \ge - 1;\)
d) \({\log _{0,5}}\left( {x + 7} \right) \ge {\log _{0,5}}\left( {2x - 1} \right).\)
\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)
\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)
c, ĐK: \(x>-7\)
\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)
Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)
d, ĐK: \(x>\dfrac{1}{2}\)
\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)
Kết hợp với ĐKXĐ, ta được: \(x\ge8\)
tim x biet
\(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)
\(\left(2x-1\right)\left(3x-1\right)-\left(3x-2\right)\left(2x-1\right)=3\)
a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy phương trình có nghiệm x = - 5 .
a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)
\(\Rightarrow3x^2-3x-3x^2+2x=5\)
\(\Rightarrow5x=5\Rightarrow x=1\)
Câu b,c làm tương tự! Cứ tách ra là làm được à!
b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)
\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)
\(\Leftrightarrow20x=-40\)
\(\Leftrightarrow x=-2\)
Vậy phương trình có nghiệm x = - 2
Tim x biet : \(\left(x-3\right)^2=\left(2x-1\right)^2\)
(x - 3)2 = (2x - 1)2
=> |x - 3| = |2x - 1|
=> x - 3 = 2x-1 hoặc x - 3 = -(2x - 1)
x - 2x = -1 + 3 hoặc x - 3 = -2x + 1
-x = 2 hoặc x + 2x = 1 + 3
x = -2 hoặc 3x = 4
x = -2 hoặc x = 4 : 3
x = -2 hoặc x = 1,(3)
1. tim x biet
\(\left|x-3\right|-\left|x+1,5\right|=0\)
Tim x, y, z biet:
\(\left(2x-1\right)^{2008}\) + \(\left(y-\dfrac{2}{5}\right)^{2008}\) + \(\left|x+y-z\right|\) = 0
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)
Cho x\(\ge\)0. Tim Min
Q=\(\frac{x^2+2x+17}{2\left(x+1\right)}\)
\(Q=\frac{x^2+2x+17}{2\left(x+1\right)}=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{x+1}{2}.\frac{8}{x+1}}=4\)
Dấu "=" tại x = 3
Tim x biet,\(\frac{0,\left(3\right)+0,\left(384615\right)+\frac{3}{13}\chi}{0,0(3)+13}=\frac{50}{85}\)