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\(2x^3+2ax-4a-4x\\ =2\left(x^3+ax-2a-2x\right)\)
câu sau thì chịu
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\(\left\{{}\begin{matrix}x+my=3\\m^2x+my=2m^2+m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\\left(m^2-1\right)x=2m^2+m-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\x=\dfrac{2m+3}{m+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m+3}{m+1}\\y=\dfrac{1}{m+1}\end{matrix}\right.\)
\(P=\left(\dfrac{2m+3}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}=\left(2+\dfrac{1}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}\)
\(=4+\dfrac{4}{m+1}+\dfrac{4}{\left(m+1\right)^2}=\left(\dfrac{2}{m+1}+1\right)^2+3\ge3\)
\(P_{min}=3\) khi \(m=-3\)
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\(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=\left(mx^2-4mx+4m\right)-\left(nx^2-4nx+4n\right)\)
\(=m\left(x^2-4x+4\right)-n\left(x^2-4x+4\right)\)
\(=\left(m-n\right)\left(x^2-4x+4\right)\)
\(=\left(m-n\right)\left(x-2\right)^2\)
\(mx^2-4mx+4m-nx^2+4nx-4n\)
\(=x^2\left(m-n\right)+4x\left(n-m\right)+4\left(m-n\right)\)
\(=x^2\left(m-n\right)-4x\left(m-n\right)+4\left(m-n\right)\)
\(=\left(x^2-4x+4\right)\left(m-n\right)\)
\(=\left(x-2\right)^2\left(m-n\right)\)
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\(\left(mx+my\right)+\left(3x+3y\right)\)
\(=m\left(x+y\right)+3\left(x+y\right)\)
\(=\left(x+y\right)\left(m+3\right)\)
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\(1.mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(m+5\right)\left(x+y\right)\)
\(2.\left(a+b\right)\left(a-b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a-b-a^2+ab-b^2\right)\)
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https://hoc24.vn/cau-hoi/tu-diem-m-nam-trong-tam-giac-abc-ke-tia-mxmymz-theo-thu-tu-vuong-goc-voi-bcacab-tren-tia-mxmymz-lan-luot-lay-cac-diem-pqr-sao-cho-mpbcmqcamrabchung-minh-rang-m-la-trong-tam-cua-tam-gia.171683942010
a, mx^2+ my - nx^2-ny
=>(mx^2+my)-(nx^2+ny)
=>m(x^2+y)-n(x^2+y)
=>(x^2+y)(m-n)
TL
Biến đổi được mx2 + tny + nx2 - ny = ( x2 + y ) ( m -n)
HT