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N
11 tháng 1 2023
\(a.\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}-2=-1\\\dfrac{4}{x}+\dfrac{3}{y}-2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b-2=-1\\4a+3b-2=5\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10}{7}\\b=\dfrac{3}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{10}{7}\Rightarrow x=\dfrac{7}{10}\\\dfrac{1}{y}=\dfrac{3}{7}\Rightarrow y=\dfrac{7}{3}\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{5}{\left(x+y\right)}=2\\\dfrac{3}{x}+\dfrac{1}{\left(x+y\right)}=\dfrac{17}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a+5b=2\\3a+b=\dfrac{17}{10}\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{x+y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\\\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow y=3\end{matrix}\right.\)
\(c.\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{1}{y+1}=7\\\dfrac{5}{x-1}-\dfrac{2}{y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\) (với \(\dfrac{1}{x-1}=a-\dfrac{1}{y+1}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=2\Rightarrow x=\dfrac{3}{2}\\\dfrac{1}{y+1}=3\Rightarrow y=-\dfrac{2}{3}\end{matrix}\right.\)
\(d.\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-1}}=1\\\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{y-1}}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\) (với \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)
11 tháng 10 2021
e: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)
11 tháng 10 2021
a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)
12 tháng 10 2021
c) \(\left\{{}\begin{matrix}2\left(x-2\right)+3\left(1+y\right)=2\\3\left(x-2\right)-2\left(1+y\right)=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left(x-2\right)+9\left(1+y\right)=6\\6\left(x-2\right)-4\left(1+y\right)=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13\left(1+y\right)=12\\2\left(x-2\right)+3\left(1+y\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21}{13}\\y=-\dfrac{1}{13}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\left(x-5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-2x-5y+10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x-7y=-12\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-x-7y=-12\\21x-7y=112\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}22x=124\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{62}{11}\\y=\dfrac{10}{11}\end{matrix}\right.\)
29 tháng 8 2021
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
T
14 tháng 1 2019
a) \(\left\{{}\begin{matrix}3x-4y=-2\\2x+y=6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}3x-4y=-2\\8x+4y=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\3x-4y=-2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
31 tháng 12 2022
a: =>3x-4y=-2 và 8x+4y=24
=>11x=22 và 2x+y=6
=>x=2 và y=6-2x=6-2*2=2
b: 2x-y=0 và 3x+y=4
=>5x=4 và y=2x
=>x=4/5 và y=8/5
c: x+3y=-2 và x-y=-1
=>4y=-1 và x=y-1
=>y=-1/4 và x=-1/4-1=-5/4
d: x+y=3 và 4x-3y=-2
=>4x+4y=12 và 4x-3y=-2
=>7y=14 và x+y=3
=>y=2 và x=1
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) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)