CC
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NN
2
14 tháng 11 2017
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-\sqrt{9}\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-3\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\left|x+\dfrac{3}{2}\right|=4,5-3,5\)
\(\left|x+\dfrac{3}{2}\right|=1\)
\(\Rightarrow x+\dfrac{3}{2}=1\) hoặc \(x+\dfrac{3}{2}=-1\)
\(x=1-\dfrac{3}{2}\) \(x=-1-\dfrac{3}{2}\)
\(x=\dfrac{-1}{2}\) \(x=\dfrac{-5}{2}\)
Vậy \(x=\dfrac{-1}{2}\)hoặc \(x=\dfrac{-5}{2}\)
TT
14 tháng 11 2017
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\)
\(\Rightarrow3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\Rightarrow\left|x+\dfrac{3}{2}\right|=4,5-3,5=1\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=1\\x+\dfrac{3}{2}=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1-\dfrac{3}{2}\\x=-1-\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
Q
18 tháng 6 2017
1) \(9^{x-1}=\dfrac{1}{9}\) (1)
\(\Leftrightarrow3^{2x-2}=3^{-2}\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{0\right\}\)
2) \(\dfrac{1}{3}:\sqrt{7-3x^2}=\dfrac{2}{15}\) (2)
\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{1}{\sqrt{7-3x^2}}=\dfrac{2}{15}\)
\(\Leftrightarrow\dfrac{1}{3\sqrt{7-3x^2}}=\dfrac{2}{15}\)
\(\Leftrightarrow15=6\sqrt{7-3x^2}\)
\(\Leftrightarrow6\sqrt{7-3x^2}=15\)
\(\Leftrightarrow\sqrt{7-3x^2}=\dfrac{5}{2}\)
\(\Leftrightarrow7-3x^2=\dfrac{25}{4}\)
\(\Leftrightarrow-3x^2=\dfrac{25}{4}-7\)
\(\Leftrightarrow-3x^2=-\dfrac{3}{4}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
\(\Leftrightarrow x=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
18 tháng 6 2017
Ta có : \(9^{x-1}=\frac{1}{9}\)
=> \(9^{x-1}=9^{-1}\)
=> x - 1 = -1
=> x = 0
ko biết bạn học mũ âm chưa nêu chưa thì mk xin lỗi
=>
29 tháng 7 2017
\(\dfrac{x-2}{2}=\dfrac{y-4}{3}=\dfrac{z-8}{5}\)
\(\Rightarrow\dfrac{x-2}{2}+2=\dfrac{y-4}{3}+2=\dfrac{z-8}{5}+2\)
\(\Rightarrow\dfrac{x+2}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{5}\)
\(\Rightarrow\left(\dfrac{x+2}{2}\right)^2=\left(\dfrac{y+2}{3}\right)^2=\left(\dfrac{z+2}{5}\right)^2\)
\(\Rightarrow\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}=\dfrac{3.\left(y+2\right)^2}{27}\dfrac{\left(x+2\right)^2+3\left(y+2\right)^2-\left(z+2\right)^2}{4+27-25}=\dfrac{24}{6}=4\)\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=16\\\left(y+2\right)^2=36\\\left(z+2\right)^2=100\end{matrix}\right.\)
Bạn chia trường hợp rồi tìm x,y,z nhé
17 tháng 6 2018
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
TM
26 tháng 2 2017
Vì \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|\ge0\);|x+y+z|\(\ge\)0
=>\(\left|x-\sqrt{2}\right|+\left|y+\sqrt{2}\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi \(\left|x-\sqrt{2}\right|=\left|y+\sqrt{2}\right|=\left|x+y+z\right|=0\)
\(\left|x-\sqrt{2}\right|=0\Leftrightarrow x-\sqrt{2}=0\Leftrightarrow x=\sqrt{2}\)
\(\left|y+\sqrt{2}\right|=0\Leftrightarrow y+\sqrt{2}=0\Leftrightarrow y=-\sqrt{2}\)
\(\left|x+y+z\right|=0\Leftrightarrow x+y+z=0\Leftrightarrow\sqrt{2}+\left(-\sqrt{2}\right)+z=0\Leftrightarrow z=0\)
Vậy ............
23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
3,5 + /x + \(\frac{3}{2}\) / = -1,5(-\(\sqrt{9}\))
=> 3,5 +/ x +\(\frac{3}{2}\) / = -1,5 ( -3 )
=> 3,5 + / x + \(\frac{3}{2}\) / =4,5
=> / x + \(\frac{3}{2}\) / = 4,5 - 3,5
=> / x + \(\frac{3}{2}\) / = 1
=> \(\hept{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\)
=> \(\hept{\begin{cases}x=1-\frac{3}{2}\\x=-1-\frac{3}{2}\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=\frac{-5}{2}\end{cases}}\)
vậy x = \(\frac{-1}{2}\)hay x = \(\frac{-5}{2}\)
\(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\) \(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-3\right)\) \(3,5+\left|x+\frac{3}{2}\right|=4,5\) \(\left|x+\frac{3}{2}\right|=4,5-3,5\) \(\left|x+\frac{3}{2}\right|=1\) \(\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\) Vậy x=\(-\frac{1}{2}\) hoặc x=\(-\frac{5}{2}\)