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\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)
\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).
a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)
b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)
a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).
b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)
\(A=\dfrac{x^{\dfrac{5}{4}}y+xy^{\dfrac{5}{4}}}{\sqrt[4]{x}+\sqrt[4]{y}}\\ =\dfrac{xy\left(x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}\right)}{x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}}\\ =xy\)
\(B=\left(\sqrt[7]{\dfrac{x}{y}\sqrt[5]{\dfrac{y}{x}}}\right)^{\dfrac{35}{4}}\\= \left(\sqrt[7]{\dfrac{x}{y}\cdot\left(\dfrac{x}{y}\right)^{-\dfrac{1}{5}}}\right)^{\dfrac{35}{4}}\\ =\left(\sqrt[7]{\left(\dfrac{x}{y}\right)^{\dfrac{4}{5}}}\right)^{\dfrac{35}{4}}\\ =\left[\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}}\right]^{\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}\cdot\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^1\\ =\dfrac{x}{y}\)
a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)
b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)
Một dạng rất uen thuộc của lượng giác là tìm gtnn,ln bằng cách đặt ẩn là sinx và cosx
\(x^2+y^2-2x-4y+4=0\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=1\)
\(\left\{{}\begin{matrix}\sin\alpha=x-1\\\cos\alpha=y-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\sin\alpha+1\\y=\cos\alpha+2\end{matrix}\right.\)
\(\Rightarrow P=\left(\sin\alpha+1\right)^2-\left(\cos\alpha+2\right)^2+2\sqrt{3}\left(\sin\alpha+1\right)\left(\cos\alpha+2\right)-2\left(\sin\alpha+1\right)-4\sqrt{3}\left(\sin\alpha+1\right)-4\left(\cos\alpha+2\right)-2\sqrt{3}\left(\cos\alpha+2\right)-3+4\sqrt{3}\)
\(\Leftrightarrow P=\sin^2\alpha-\cos^2\alpha+2\sqrt{3}\sin\alpha\cos\alpha-16\)
Ta đưa về góc 2 alpha để dễ xét
\(\Leftrightarrow P=\frac{1-\cos2\alpha}{2}-\frac{\cos2\alpha+1}{2}+\sqrt{3}\sin2\alpha-16\)
\(\Rightarrow P=\sqrt{3}\sin2\alpha-\cos2\alpha-16\)
\(P=2\sin\left(2\alpha-\frac{\pi}{6}\right)-16\)
\(\Rightarrow2.\left(-1\right)-16\le P\le2.1-16\)
\(\Rightarrow\left\{{}\begin{matrix}P_{min}=-18;"="\Leftrightarrow2\alpha-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\\P_{max}=-14;"="\Leftrightarrow2\alpha-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
Bạn tự thay vô x và y để xét dấu bằng nhé
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
\(N=\dfrac{xy\left(x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}\right)}{x^{\dfrac{1}{3}}+y^{\dfrac{1}{3}}}=xy\)