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Vì \(\left(9x^2-1\right)^2\ge0;\left|x-\frac{1}{3}\right|\ge0\Rightarrow\left(9x^2-1\right)^2+\left|x-\frac{1}{3}\right|\ge0\)
Để \(\left(9x^2-1\right)^2+\left|x-\frac{1}{3}\right|=0\Leftrightarrow\hept{\begin{cases}9x^2-1=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow x=\frac{1}{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ............
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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
=>
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\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\\x=25\end{matrix}\right.\)
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a/ \(\Leftrightarrow9x^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow x=\pm2\)
b/ \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\) (do \(x^2+\dfrac{1}{2}>0\))
\(\Leftrightarrow x=\pm1\)
c/ Có \(\left|x+4\right|\ge0\forall x\)
=> \(\left|x+4\right|+5\ge5>0\forall x\)
\(\Rightarrow\left|x+4\right|+5=0\left(vô-lí\right)\)
\(\Rightarrow x\in\varnothing\)
d/ \(\sqrt{2x}-3-1=0\)
\(\Leftrightarrow\sqrt{2x}=4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
Vì có mũ 2 nên được bỏ căn. Cái này lớp 9 mới dạy :((
(=) 2 - 9x + 2 ( 1 + x ) = 0
(=) 2 - 9x + 2 + 2x = 0
(=) - 9x + 2x = 0 - 2 - 2
(=) - 7x = -4
(=) x = -4 : ( -7 )
(=) x = \(\frac{4}{7}\)