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a, \(4y^2+1-4y=\left(2y\right)^2-2.2y.1+1^2=\left(2y-1\right)^2\)
b, \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)
c, \(x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
D= 5x^2+8xy+5y^2-2x+2y
=4x^2+8xy+4y^2-2x+2y+y^2+x^2
=(2x+2y)^2+x^2-2*1/2x+1/4+y^2+2*1/2y+1/4-1/2
(2x+2y)^2+(x-1/2)^2+(y+1/2)^2-1/2>=-1/2
suy ra D>=-1/2 nên D có GTNN là -1/2
Ta có : 5D = 25x2 + 40xy + 25y2 - 10x + 10y
5D = (5x+ 4y - 1)2 + 9y2 + 18y - 1
5D = ( 5x + 4y - 1)2 + 9 (y + 1)2 - 2
D =\(\frac{1}{5}\). ( 5x + 4y - 1)2 + \(\frac{9}{5}\).( y + 1)2 - \(\frac{2}{5}\) \(\ge\)\(\frac{-2}{5}\)
Dấu "=" xảy ra khi y+1 = 0 \(\Leftrightarrow\)y = -1
5x + 4y - 1 = 0 \(\Leftrightarrow\)x=1
Vậy GTNN của D = \(\frac{-2}{5}\)khi x = 1 ; y = -1
áp dụng công thức này mà lm câu a,b,e nhá:
\(A=ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}\\ \left[{}\begin{matrix}A\ge\dfrac{4ac-b^2}{4a}\left(với\text{ }\text{ }\text{ }a\ge0\right)\\A\le\dfrac{4ac-b^2}{4a}\left(với\text{ }a< 0\right)\end{matrix}\right.\)
\(C=x^2+2xy+y^2+4y^2=\left(x+y\right)^2+4y^2\ge0\)
đẳng thức xảy ra khi x=y=0
vậy MIN C=0 tại x=y=0
\(A=x^2-xy+\frac{y^2}{4}+\frac{3}{4}\left(y^2-4y+4\right)+2013\)
\(=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}\left(y-2\right)^2+2013\ge2013\)
\(B\) đề thiếu
\(C\) đề sai, dấu của \(y^2\) là âm thì không tồn tại GTNN
\(P=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
\(2Q=-4x^2-20y^2+12xy+8x-6y+4\)
\(=-\left(4x^2+9y^2+4-12xy-8x+12y\right)-11\left(y^2-\frac{6}{11}y+\frac{36}{121}\right)+\frac{97}{11}\)
\(=-\left(2x-3y-2\right)^2-11\left(y-\frac{3}{11}\right)^2+\frac{97}{11}\le\frac{97}{11}\)
\(\Rightarrow Q\le\frac{97}{22}\)
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
Những câu khác làm tương tự
\(a)\)
\(A=2x^2+x\)
\(\Leftrightarrow A=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
\(MinA=\frac{-1}{8}\)khi \(x=\frac{-1}{4}\)
\(b)\)
\(B=x^2+2x+y^2-4y+6\)
\(\Leftrightarrow B=x^2+2x+1+y^2-4y+4+1\)
\(\Leftrightarrow B=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu '' = '' xảy ra khi: \(x=-1;y=2\)
\(c)\)
\(C=4x^2+4x+9y^2-6y-5\)
\(\Leftrightarrow C=4x^2+4x+1+9y^2-6y+1-7\)
\(\Leftrightarrow C=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
Dấu '' = '' xáy ra khi: \(x=\frac{-1}{2};y=\frac{1}{3}\)
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
a) A = x2 - 4y2 + 2x + 4y = (x-2y)(x+2y)+2(x+2y)=(x+2y)(x-2y+2)
b) A = 4x2 - 9y2 - 4x - 6y=(2x-3y)(2x+3y)-2(2x+3y)=(2x+3y)(2x-3y-2)
c) A = 3x2 - 3xy - 5x + 5y=3x(x-y)-5(x-y)=(x-y)(3x-5)
a) \(A=x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
b) \(A=4x^2-9y^2-4x-6y=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)=\left(2x+3y\right)\left(2x-3y-2\right)\)
c) \(A=3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)