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a)
\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Daaus = xayr ra khi: x = 2
b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)
Dấu = xảy ra khi x = 3
c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu = xảy ra khi
2x = y và y = 2
=> x = 1 và y = 2
a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)
Dấu "=" <=> x = 2
b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)
Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)
c) \(4x^2+2y^2-4xy-4y+1\)
= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)
= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Đặt \(x+2=t\ne0\Rightarrow x+1=t-1\)
\(A=\dfrac{x+1}{\left(x+2\right)^2}=\dfrac{t-1}{t^2}=-\dfrac{1}{t^2}+\dfrac{1}{t}=-\left(\dfrac{1}{t}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(A_{max}=\dfrac{1}{4}\) khi \(t=2\) hay \(x=0\)
\(B=\left(x-1\right)^2-4\ge4\\ B_{min}=4\Leftrightarrow x=1\)
\(B=x^2-2x-3=\left(x^2-2x+1\right)-4\)
\(=\left(x-1\right)^2-4\ge-4\)
\(minB=-4\Leftrightarrow x=1\)
\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5\)
Ta thấy \(5x^2\ge0\forall x\)
\(\Rightarrow5x^2+5\ge5\)
\(\Rightarrow B\ge5\)
Dấu "=" xảy ra khi \(x=0\)
...
\(B=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5\ge5\)
Dấu "=" xảy ra <=> x^2 = 0 <=> x = 0
GTNN của B là 5 khi x = 0
4-\(x^2\)+2x
=-x\(^2\)+2x-1+5
=-(x\(^2\)-2x+1)+5
=-(x-1)\(^2\)+5
có(x-1)\(^2\)\(\ge\)0\(\forall\)x\(\in\)R
=>-(x-1)\(^2\)\(\le\)0\(\forall\)x\(\in\)R
=>-(x-1)\(^2\)+5\(\le\)5\(\forall\)x\(\in\)R
vậy GTLN của bt trên là 5 \(\Leftrightarrow\)x=1
\(a,f\left(x\right)⋮g\left(x\right)\\ \Leftrightarrow\dfrac{-x^4+2x^2-3x+5}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^4+x^3-x^3+x^2+x^2-x-2x+2+3}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^3\left(x-1\right)-x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)+3}{x-1}\in Z\\ \Leftrightarrow-x^3-x^2+x-2+\dfrac{3}{x-1}\in Z\\ \Leftrightarrow3⋮x-1\\ \Leftrightarrow x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\\ Mà.x< 0\\ \Leftrightarrow x=-2\\ b,B=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y\right)^2+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y-2\right)^2+4y^2-2024\ge-2024\\ B_{min}=-2024\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
\(A=-x^2+3x-7\)
\(=-\left(x^2-3x+7\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{19}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}< 0\forall x\)
\(3x-7-x^2=-\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{19}{4}=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}< 0\)
x - x^2 = -(x^2 - x ) = - ( x^2 - x + 1/4 ) + 1/4 = - (x-1/2)^2 + 1/4
Mà - (x-1/2)^2 \(\le\)0 ''='' <=> x = 1/2
=> Max B = 1/4 <=> x =1/2