Câu 6:
\(\left(\dfrac{3-2x}{\sqrt{4x-1}}\right)'=\dfrac{\left(3-2x\right)'\cdot\sqrt{4x-1}-\left(3-2x\right)\cdot\left(\sqrt{4x-1}\right)'}{4x-1}\)
\(=\dfrac{-2\sqrt{4x-1}-\left(3-2x\right)\cdot\dfrac{\left(4x-1\right)'}{2\sqrt{4x-1}}}{4x-1}\)
\(=\dfrac{-2\sqrt{4x-1}-\left(3-2x\right)\cdot\dfrac{2}{\sqrt{4x-1}}}{4x-1}\)
\(=\dfrac{-2\left(4x-1\right)-\left(3-2x\right)\cdot2}{\sqrt{4x-1}\cdot\left(4x-1\right)}\)
\(=\dfrac{-8x+2-6+4x}{\left(\sqrt{4x-1}\right)\left(4x-1\right)}=\dfrac{-4x-4}{\sqrt{4x-1}\cdot\left(4x-1\right)}\)
=>a=-4; b=4
=>a/b=-1
Câu 4:
\(\widehat{SD;\left(ABCD\right)}=\widehat{DS;DA}=\widehat{SDA}\)
Xét ΔSAD vuông tại A có \(tanSDA=\dfrac{SA}{AD}=\sqrt{3}\)
nên \(\widehat{SDA}=60^0\)
=>\(\widehat{SD;\left(ABCD\right)}=60^0\)
Câu 3:
\(y=-\dfrac{3}{2}x^2+\dfrac{2}{x}+\dfrac{x^3}{3}\)
=>\(y'=\dfrac{-3}{2}\cdot2x+\dfrac{2'\cdot x-2\cdot x'}{x^2}+\dfrac{1}{3}\cdot3x^2\)
=>\(y'=-3x-\dfrac{2}{x^2}+x^2\)
\(y'\left(2\right)=-3\cdot2-\dfrac{2}{2^2}+2^2=-2-\dfrac{2}{4}=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
Câu 2: ĐKXĐ: x>-4
\(log_2\left(x+4\right)=3\)
=>\(x+4=2^3=8\)
=>x=4(nhận)
Câu 1: ĐKXĐ: 1-5x>0
=>5x<1
=>x<1/5
=>a=1/5
