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\(log_{a^2}\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}log_a\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}\left[log_aa^3-log_a\sqrt[5]{b^3}\right]=\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)\)
\(\Rightarrow\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)=3\)
\(\Rightarrow log_ab=-5\)
\(\int\sqrt{\dfrac{2+x}{2-x}}dx=\int\dfrac{2+x}{\sqrt{4-x^2}}dx\)
Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)
\(\Rightarrow I=\int\dfrac{2+2sinu}{\sqrt{4-4sin^2u}}.2cosu.du=\int\left(2+2sinu\right)du=2u-2cosu+C\)
Trả biến: \(sinu=\dfrac{x}{2}\Rightarrow u=arcsin\left(\dfrac{x}{2}\right)\)
\(x=2sinu\Rightarrow x^2=4sin^2u=4-4cos^2u\Rightarrow cos^2u=1-\dfrac{x^2}{4}\)
\(\Rightarrow cosu=\sqrt{1-\dfrac{x^2}{4}}\)
Vậy \(I=2arcsin\left(\dfrac{x}{2}\right)-\sqrt{4-x^2}+C\)
\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)
\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)
\(54=7q+r\)
\(\Rightarrow54-r=7q\)
Mà \(0\le r< 7\)
\(\Rightarrow7q\in\left\{54;53;52;51;50;49;48\right\}\)
\(\Rightarrow q\in\left\{\dfrac{54}{7};\dfrac{53}{7};\dfrac{52}{7};\dfrac{51}{7};\dfrac{50}{7};7;\dfrac{48}{7}\right\}\)
Mà \(q\in N\)
\(\Rightarrow q=7\Rightarrow r=54-7q=54-7.7=5\)
Vậy \(q=7;r=5\)
x3 và 3x2