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\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3x-6}\)
\(\Leftrightarrow x=3x-6\)
\(\Leftrightarrow3x-x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ........
\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1^3}{2^3}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3\left(x-2\right)}\\ \Leftrightarrow3\left(x-2\right)=x\\ \Rightarrow3x-6=x\\ \Rightarrow3x-x=6\\ \Rightarrow x\left(3-1\right)=6\\ \Rightarrow2x=6\\ \Rightarrow x=6:2=3\)
\(\left(2+4x\right)^2+\left(y-6\right)^2=0\)
\(\left\{{}\begin{matrix}\left(2+4x\right)^2\ge0\\\left(y-6\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(2+4x\right)^2+\left(y-6\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(2+4x\right)^2=0\Rightarrow2+4x=0\Rightarrow4x=-2\Rightarrow x=-0,5\\\left(y-6\right)^2=0\Rightarrow y-6=0\Rightarrow y=6\end{matrix}\right.\)
\(\left|8-4x\right|+\left|2x-y\right|=0\)
\(\left\{{}\begin{matrix}\left|8-4x\right|\ge0\\\left|2x-y\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|8-4x\right|+\left|2x-y\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|8-4x\right|=0\Rightarrow8-4x=0\Rightarrow4x=8\Rightarrow x=2\\2.2-y=0\Rightarrow y=4\end{matrix}\right.\)
\(\left|16+0,5x\right|+\left(y-2\right)^2=0\)
\(\left\{{}\begin{matrix}\left|16+0,5x\right|\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left|16+0,5x\right|+\left(y-2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|16+0,5x\right|=0\Rightarrow16+0,5x=0\Rightarrow0,5x=16\Rightarrow x=32\\\left(y-2\right)^2=0\Rightarrow y-2=0\Rightarrow y=2\end{matrix}\right.\)
x2 + 2x = 0
=> x(x + 2) = 0
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
(x - 2) + 3.x2 - 6x = 0
=> (x - 2) + 3x2 - 3x . 2 = 0
=> (x - 2) + 3x.(x - 2) = 0
=> (1 + 3x)(x - 2) = 0
=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
a) \(\frac{4}{3}-\frac{2}{5}\)
\(=\frac{20}{15}-\frac{6}{15}=\frac{14}{15}\)
b) \(\left|-\frac{1}{10}\right|-\left(-\frac{1}{3}\right)^2\div\frac{5}{9}\)
\(=\frac{1}{10}-\frac{1}{9}\cdot\frac{9}{5}\)
\(=\frac{1}{10}-\frac{1}{5}=\frac{1}{10}-\frac{2}{10}\)
\(=-\frac{1}{10}\)
c) Đề bài có vấn đề!!!
d) \(\left(-0,2\right)^2\cdot5-8^2\cdot\frac{9^4}{3^7}\cdot4^3\)
\(=0,04\cdot5-64\cdot\frac{\left(3^2\right)^4}{3^7}\cdot64\)
\(=0,2-4096\cdot\frac{3^8}{3^7}=0,2-4096\cdot3\)
\(=0,2-12288=-128878\)
\(\Rightarrow\left(3x+2\right).\left(5x-3\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Rightarrow15x^2-9x+10x-6=15x^2-5x+21x-7\)
\(\Rightarrow19x-6=26x-7\)
\(\Rightarrow26x-19x=7-6\)
\(\Rightarrow13x=1\)
\(\Rightarrow x=\frac{1}{13}\)
\(\left|x-3\right|=3x-2\Leftrightarrow\hept{\begin{cases}x-3=3x-2\\x-3=-\left(3x-2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}4x=1\\x-3=2-3x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\4x=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\x=\frac{5}{4}\end{cases}}}\)
ta có x2y + xy - x = xy (x+1)-x-1=xy(x+1) - (x+1) = (x+1)(xy-1)=5
( 3x - 2 )2 + 1,25 = 7,5
=> ( 3x - 2 )2 = 6,25
\(\Rightarrow\left(3x-2\right)^2=\frac{25}{4}\)
\(\Rightarrow\left(3x-2\right)^2=\left(\frac{5}{2}\right)^2=\left(\frac{-5}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x-2=\frac{5}{2}\\3x-2=\frac{-5}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-2=\frac{5}{2}\\3x-2=\frac{-5}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=\frac{9}{2}\\3x=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-1}{6}\end{cases}}\)