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\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(\Leftrightarrow A=\left(x^2-x+6x-6\right)\left(x^2+2x+3x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(\Leftrightarrow A=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy GTNN của A là : \(-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
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B= ( y^4 - 4y^2 + 4 ) + ( x^2 + 4xy + 4 ) +2025
B= ( y^2 - 2 )^2 + ( x + 2 )^2 + 2025 >hoặc=2025
=> minB = 2025
chỉ vậy thôi dạng này áp dụng hđt rùi tách ra là ok bn nhé =))
B= ( y^4 - 4y^2 + 4 ) + ( x^2 + 4xy + 4 ) +2025
B= ( y^2 - 2 )^2 + ( x + 2 )^2 + 2025 >hoặc=2025
=> minB = 2025
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Áp dụng \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
Ta có \(P=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)^3-3x^2y^2\left(x^2+y^2\right)\)
\(\Rightarrow P=1-3x^2y^2\ge1-3\dfrac{\left(x^2+y^2\right)^2}{4}=\dfrac{1}{4}\)
\(\Rightarrow P_{min}=\dfrac{1}{4}\) khi \(x^2=y^2=\dfrac{1}{2}\)
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Đặt \(A=x^2+y^2-x+6y+10\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0;\)\(\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}\)
Dấu"=" xảy ra khi \(\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2};\left(y-3\right)^2=0\Leftrightarrow y=3\)
Vậy \(MinA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=3\)
Đặt x = 4 - m; y = 4 + m
=> x2 + y2 = (4 - m)2 + (4 + m)2 = 16 - 8m + m2 + 16 + 8m + m2 = 32 + 2m2
Vì m2 >= 0 => 2m2 >= 0
=> 32 + 2m2 >= 32
Dấu bằng xảy ra khi: m2 = 0 => m = 0
Vậy x2 + y2min = 32 <=> x = y = 4
Ta có: \(x+y=4\) \(\Rightarrow\) \(y=4-x\)
Do đó: \(A=x^2+y^2=x^2+\left(4-x\right)^2=x^2+16-8x+x^2=2x^2-8x+16=2\left(x^2-4x+4\right)+8\)
\(A=2\left(x-2\right)^2+8\ge8\) với mọi \(x;y\)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\left(x-2\right)^2=0\)
\(\Leftrightarrow\) \(x-2=0\)
\(\Leftrightarrow\) \(x=2\)
\(\Rightarrow\) \(y=2\) (do \(x+y=4\) )
Vậy, \(Min\) \(A=8\) \(\Leftrightarrow\) \(x=y=2\)