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Độ PH của mẫu 1 là:
\(a=-log\left[H^+\right]=-log\left[8\cdot10^{-7}\right]=-\left(log8-7\right)\)
\(=7-log8=7-log2^3=7-3\cdot log2\)
Độ PH của mẫu 2 là:
\(b=-log\left[2\cdot10^{-9}\right]=-\left(log2-9\right)=9-log2\)
\(a-b=7-3\cdot log2-9+log2=-2log2-2< 0\)
=>a<b
=>Độ PH của mẫu 2 lớn hơn
Ta có: \(lim\dfrac{3-2x}{\sqrt{x}-3}=lim\dfrac{\dfrac{3}{x}-2}{\dfrac{1}{\sqrt{x}}-\dfrac{3}{x}}=-\infty\)
Vì: \(lim\left(\dfrac{3}{x}-2\right)=-2< 0\)
\(lim\left(\dfrac{1}{\sqrt{x}}-\dfrac{3}{x}\right)=0\) và \(\dfrac{1}{\sqrt{x}}-\dfrac{3}{x}>0\) khi x vô cùng lớn.
http://123doc.org/document/1883740-phuong-phap-dung-truc-toa-do-trong-bai-hinh-hoc-khong-gian-new.htm
cả nhà giúp mình với mai minh kiểm tra chất lượng rồi. Thanks all.
Đặt \(x+\dfrac{1}{x}=t\Rightarrow t^2=x^2+\dfrac{1}{x^2}+2\)
Pt trở thành:
\(7t+2\left(t^2-2\right)=5\Leftrightarrow2t^2+7t-9=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vô-nghiệm\right)\\x^2+\dfrac{9}{2}x+1=0\end{matrix}\right.\)
Theo hệ thức Viet: \(x_1x_2=\dfrac{c}{a}=1\)
a) \(\sin^4x=\left(\sin^2x\right)^2=\left(\dfrac{1-\cos2x}{2}\right)^2\)
\(=\dfrac{1}{4}\left(1-2\cos2x+\cos^22x\right)\)
\(=\dfrac{1}{4}\left(1-2.\cos2x+\dfrac{1+\cos4x}{2}\right)\)
\(=\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\)
Vậy:
\(\int\sin^4x\text{dx}=\int\left(\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\right)\text{dx}\)
\(=\dfrac{3}{8}x-\dfrac{1}{4}\sin2x+\dfrac{1}{32}\sin4x+C\)
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999x1009999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 = 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999x1009999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
Khi góc α rất nhỏ (nhỏ hơn 10 độ) thì \(sin\left(x\right)\approx x\) nên \(sin\left(1^o\right)\approx\dfrac{\pi}{180}\approx0.01745\)