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b: Ta có: \(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
\(\Leftrightarrow\dfrac{3}{5}-\dfrac{2}{3}x=\dfrac{-4}{5}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}\)
hay \(x=\dfrac{7}{5}:\dfrac{2}{3}=\dfrac{21}{10}\)
a) Do \(\left(3x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(3x-\dfrac{1}{2}\right)^2-4\ge-4\)
\(minA=-4\Leftrightarrow x=\dfrac{1}{6}\)
b) Do \(\left(2x+1\right)^4\ge0\forall x,\left(y-\dfrac{1}{2}\right)^6\ge0\forall y\)
\(\Rightarrow B=\left(2x+1\right)^4+3\left(y-\dfrac{1}{2}\right)^6\ge0\)
\(minB=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
a: \(A=\left(3x-\dfrac{1}{2}\right)^2-4\ge-4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{6}\)
b: \(B=\left(2x+1\right)^4+3\left(y-\dfrac{1}{2}\right)^6\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-\dfrac{1}{2};\dfrac{1}{2}\right)\)
Đề:........
<=> (24)x < (27)4
<=> 24x < 228
<=> 4x < 28
<=> x < 7
Vậy x = {0; 1; 2; 3; 4; 5; 6}
\(\left(x-3\right)^{30}=\left(x-3\right)^{10}\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)
hơi tắt ạ