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\(M=\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ M_{min}=-\dfrac{1}{4}\Leftrightarrow x=y=-\dfrac{1}{2}\)
\(a,A=4-x^2+2x=4-\left(x^2-2x\right)=4-\left(x^2-2x+1-1\right)\)
\(=4-\left[\left(x-1\right)^2-1\right]=4-\left(x-1\right)^2+1=5-\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0=>-\left(x-1\right)^2\le0=>5-\left(x-1\right)^2\le5\) (với mọi x)
Dấu "=" xảy ra \(< =>\left(x-1\right)^2=0< =>x=1\)
Vậy MaxA=5 khi x=1
\(b,B=4x-x^2=-x^2+4x=-\left(x^2-4x\right)=-\left(x^2-4x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4=4-\left(x-2\right)^2\)
Vì \(\left(x-2\right)^2\ge0=>-\left(x-2\right)^2\le0=>4-\left(x-2\right)^2\le4\) (với mọi x)
Dấu "=" xảy ra \(< =>\left(x-2\right)^2=0< =>x=2\)
Vậy MaxB=4 khi x=2
a) \(4-x^2+2x\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(\left(x-1\right)^2-5\right)\)
\(=5-\left(x-1\right)^2\ge5\)
MIn A = 5 khi \(x-1=0=>x=1\)
b) \(4x-x^2\)
\(=-\left(x^2-4x+4-4\right)\)
\(=>-\left(\left(x-2\right)^2-4\right)\)
\(=4-\left(x-2\right)\ge4\)
MIN B = 4 khi \(x-2=0=>x=2\)
Ủng hộ nha tối rồi
đặt y = 1/x suy ra y <=1,
ta có P = 1 -2y+2016y^2
Tự làm tiếp nhé
Vừa học xong :v
\(A=\frac{4}{4x^2-4x+7}\)
Ta có : \(4x^2-4x+7=4x^2-4x+1+6\)
\(=\left(2x-1\right)^2+6\ge6\)Do đó :
\(\frac{4}{\left(2x-1\right)^2+6}\le\frac{4}{6}=\frac{2}{3}\)
Dấu ''='' xảy ra : <=> \(x=\frac{1}{2}\)
Vậy GTLN A = 2/3 <=> x = 1/2
Ta có : 4x2 - 4x + 7
= ( 4x2 - 4x + 1 ) + 6
= ( 2x - 1 )2 + 6 ≥ 6 ∀ x
hay 4x2 - 4x + 7 ≥ 6 ∀ x
=> \(\frac{1}{4x^2-4x+7}\le\frac{1}{6}\left(\forall x\right)\)
=> \(\frac{4}{4x^2-4x+7}\le\frac{4}{6}=\frac{2}{3}\left(\forall x\right)\)
Đẳng thức xảy ra khi x = 1/2
=> MaxA = 2/3 <=> x = 1/2
A=−2x2−10y2+4xy+4x+4y+2016A=−2x2−10y2+4xy+4x+4y+2016
=−2.(x2+5y2−4xy−4x−4y)+2016=−2.(x2+5y2−4xy−4x−4y)+2016
=−2.(x2+4y2+4−4xy−4x+8y+y2−12y+36)+2.36+2016=−2.(x2+4y2+4−4xy−4x+8y+y2−12y+36)+2.36+2016
=−2.[(x−2y−2)2+(y−6)2]+2088=−2.[(x−2y−2)2+(y−6)2]+2088
Ta có: (x−2y−2)2+(y−6)2≥0(x−2y−2)2+(y−6)2≥0
⇒−2.[(x−2y−2)2+(y−6)2]≤0⇒−2.[(x−2y−2)2+(y−6)2]≤0
⇒−2.[(x−2y−2)2+(y−6)2]+2088≤2088⇒−2.[(x−2y−2)2+(y−6)2]+2088≤2088
⇒A≤2088⇒A≤2088
Vậy giá trị lớn nhất của A=2088A=2088 khi: \hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6\hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6
Thu gọn
\(A=-2\left(x^2+2xy+y^2\right)+4\left(x+y\right)-2-8y^2+2018\\ A=-2\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-8y^2+2018\\ A=-2\left(x+y-1\right)^2-8y^2+2018\le2018\\ A_{max}=2018\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
C = 2x^2 + y^2 + 2xy - 4x - 2016
C = (x^2 + 2xy + y^2) + (x^2 - 4x + 4) - 2020
C = (x + y)^2 + (x - 2)^2 - 2020
(x+y)^2 > 0; (x - 2)^2 > 0
C > -2020
dấu "=" xảy ra khi x + y = 0 và x - 2 = 0
<=> x = 2; y = -2
\(x^2+2\ge2\Rightarrow\frac{6}{x^2+2}\le\frac{6}{2}=3\)
Vay Max D=3, dau = xay ra khi x=0