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`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):}`
`B=x^2+y^2-2x+4y+2010`
`B=x^2-2x+y^2+4y+2010`
`B= x^2-2.x.1+1^2-1^2 +y^2+2y.2+2^2-2^2+2010`
`B= (x^2-2x+1)+(y^2+4y+4)-1-4+2010`
`B= (x-1)^2 +(y+2)^2 +2005≥2005`
nên `B` đạt GTNN là `B=2005`
khi đó \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\) `<=>`\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
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Ta có : \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left(2x-1\right)^2-4\le-4\)
\(\Rightarrow B\ge\dfrac{15}{-4}\)
Dấu ''='' xảy ra khi x = 1/2
Vậy GTNN B là -15/4 khi x = 1/2
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Đề bài ko chính xác
Biểu thức này chỉ có GTLN, không có GTNN
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Ta có: \(\left(-2x+1\right)\left(x+3\right)+\left(x+1\right)\left(2x-1\right)=14\)
\(\Leftrightarrow-2x^2-6x+x+3+2x^2-x+2x-1=14\)
\(\Leftrightarrow-4x=12\)
hay x=-3
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\(\left(x-1\right)^2=\left(2x+14\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x+14\\x-1=-\left(2x+14\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-14\\x-1=-2x-14\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\x+2x=-14+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\3x=-13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=-\dfrac{13}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-15;-\dfrac{13}{3}\right\}\)
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Câu 9:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)
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A(x)=(-3x² - 2x - 14) (-2x²) - x( 2x² + 3x - 2)=6x^4 +4x^3 +28x^2 -2x^3 -3x^2 +2x= 6x^4 +2x^3 +25x^2 +2x
vậy hệ số của x^3 là 2, hệ số của x^2 là 25
B=2(x^2+2.x.1/4 +1/16)^2 -57/8
=2.(x+1/4)^2 -57/8
MinB=-57/8 khi x=-1/4
\(B=-14+2x^2+x=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{113}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{113}{8}\ge-\dfrac{113}{8}\)\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)