HOC24
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Chứng minh rằng nếu \(\left(1-sinA\right)\left(1-sinB\right)=cosA.cosB\) thì \(\left(1+sinA\right)\left(1+sinB\right)=cosA.cosB\)
`2,25 xx 0,5 + 2,25 : 2 + 3,75`
`=2,25 xx 0,5 + 2,25 xx 1/2 + 3,75`
`=2,25 xx 0,5 + 2,25 xx 0,5 + 3,75`
`=2,25 xx (0,5 + 0,5) + 3,75`
`=2,25 xx 1 +3,75`
`=2,25+3,75`
`=6`
\(\dfrac{3x-5}{8}-\dfrac{5x-4}{6}-\dfrac{6x+1}{10}=1\\ \Leftrightarrow\dfrac{15\left(3x-5\right)}{8.15}-\dfrac{20\left(5x-4\right)}{6.20}-\dfrac{12\left(6x+1\right)}{10.12}=1\\ \Leftrightarrow\dfrac{45x-75}{120}-\dfrac{100x-80}{120}-\dfrac{72x+12}{120}=1\\ \Leftrightarrow\dfrac{45x-75-100x+80-72x-12}{120}=1\\ \Leftrightarrow-127x-7=120\\ \Leftrightarrow-127x=127\\ \Leftrightarrow x=-1\)
Vậy `S={-1}`
\(\left\{{}\begin{matrix}2x+y=-1\\x-2y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x=5\\x-2y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\1-2y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
Vậy `(x,y)=(1,-3)`
`B=x^2 +y^2 -2x+4y+2010`
`=x^2 -2x+1+y^2 +4y+4+2005`
`=(x-1)^2 + (y+2)^2 +2005 >= 2005`
Dấu "=" xảy ra `<=>{(x-1=0),(y+2=0):}<=>{(x=1),(y=-2):}`
Vậy `B_(min) = 2005 <=> {(x=1),(y=-2):}`