Tính tổng:\(\dfrac{5}{2x^2\left(6x+y\right)}\)+\(\dfrac{3}{5xy\left(6x+y\right)}\)
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\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)
\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)
\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)
Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)
\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)
\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)
\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)
\(\Leftrightarrow4x+y\le5\)
\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)
\(P_{max}=1\) khi \(x=y=1\)
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a. Làm gọn 1 chút xíu:
\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)
\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)
b.
\(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)
c.
\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)
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a)
Ta có: $2x^2+2y^2=5xy \Leftrightarrow 2\frac{x}{y}+\frac{y}{x}=5$
Đặt $t=\frac{x}{y}$, ta có $2t+\frac{1}{t}=5 \Rightarrow 2t^2-5t+1=0$
Giải phương trình trên ta được $t_1=\frac{1}{2}$ và $t_2=1$. Vì $0<x<y$ nên $t>0$, do đó $t=\frac{x}{y}=\frac{1}{2}$.
Từ đó suy ra $x=\frac{y}{2}$ và thay vào biểu thức $E$ ta được:
$E=\frac{x^2+y^2}{x^2-y^2}=\frac{\frac{y^2}{4}+y^2}{\frac{y^2}{4}-y^2}=-\frac{5}{3}$
Vậy kết quả là $E=-\frac{5}{3}$.
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\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
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a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
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ĐKXĐ: \(x,y\ne0\)
Hệ pt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6x+6y}{xy}=5\\\dfrac{4}{x}+\dfrac{3}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+\dfrac{6}{y}=5\\\dfrac{4}{x}+\dfrac{3}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+\dfrac{6}{y}=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+2.\left(1-\dfrac{4}{x}\right)=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+2-\dfrac{8}{x}=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2}{x}=3\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{3}{7}\end{matrix}\right.\)(tm)
Vậy...
\(\dfrac{5}{2x^2\left(6x+y\right)}+\dfrac{3}{5xy\left(6x+y\right)}\)
\(=\dfrac{5\cdot5y}{2x^2\left(6x+y\right)\cdot5y}+\dfrac{3\cdot2x}{5xy\left(6x+y\right)\cdot2x}\)
\(=\dfrac{25y}{10x^2y\left(6x+y\right)}+\dfrac{6x}{10x^2y\left(6x+y\right)}\)
\(=\dfrac{25y+6x}{10x^2y\left(6x+y\right)}\)