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1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)
2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)
3:
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)
4:
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)
3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)
\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)
\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=-\sqrt{a}\)
4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
1) Gọi điểm cố định là \(M\left(x_0;y_0\right)\)
\(\Leftrightarrow mx_0-m+1=y_0\) \(\left(\forall m\right)\)
\(\Leftrightarrow m\left(x_0-1\right)=y_0-1\) \(\left(\forall m\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_0-1=0\\y_0-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_0=1\\y_0=1\end{matrix}\right.\)
Vậy (d1) luôn đi qua điểm cố định \(\left(1;1\right)\)
2) Xét phương trình hoành độ giao điểm của (d2) và (d3)
\(2x+3=x+1\) \(\Leftrightarrow x=-2\), thay vào (d3) ta được \(y=-1\)
\(\Rightarrow\) (d3) cắt (d2) tại \(F\left(-2;-1\right)\)
Để 3 đường cắt nhau tại 1 điểm \(\Leftrightarrow F\in\left(d_1\right)\)
\(\Leftrightarrow-2m-m+1=-1\) \(\Leftrightarrow m=\dfrac{2}{3}\)
Vậy ...
Ct tổng quát:n.n!=(n+1-1)n!=(n+1)n!-1.n!=(n+1)!-n!.Sau đó thay vào >>>A=19!-1
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{2}{x-y}+\frac{6}{x+y}=1,1\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x-y}+\frac{12}{x+y}=2,2\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\frac{21}{x+y}=2,1\\\frac{2}{x-y}=1,1-\frac{6}{x+y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\\frac{2}{x-y}=1,1-\frac{6}{x+y}=1,1-\frac{6}{10}=\frac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)(tm)
Vậy hệ phương trình có nghiệm duy nhất là (x;y) = (7;3)
another way to solve
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1,1-6b}{2}\\\frac{4\cdot\left(1,1-6b\right)}{2}-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{1}{10}\\a=\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-y}=\frac{1}{4}\\\frac{1}{x+y}=\frac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x+y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)
Vậy....
2 bạn à
=2 k cho mk