Tìm GTNN :50x2+45y2+4-54xy+30x-24y
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e, (30x3+120x)-(18x2y+72y)
=30x(x2+4)-18y(x2+4)
=(x2+4)(30x-18y)
f,(70x+20xy)-(84y+24y2)
= 10x(7+2y)-12y(7+2y)
=)7+2y)(10x-12y)
e) 30x3 - 18x2y - 72y +120x
= 6 ( 5x3 - 3x2y - 12y + 20x )
= 6 ( x2 (5x - 3y) + 4 (5x - 3y)
= 6 (x2+4)(5x-3y)
\(30x^3-18x^2y-72y+120x\)
= \(6\left(5x^3-3x^2y-12y+20x\right)\)
= \(6\left[x^2\left(5x-3y\right)+4\left(5x-3y\right)\right]\)
= \(6\left(x^2+4\right)\left(5x-3y\right)\)
\(70x-84y+20xy-24y^2\)
= \(2\left(35x-42y+10xy-12y^2\right)\)
= \(2\left[7\left(5x-6y\right)+2y\left(5x-6y\right)\right]\)
= \(2\left(5x-6y\right)\left(7+2y\right)\)
\(B=\left|5x-2\right|+\left|5x-3\right|\)
\(=\left|5x-2\right|+\left|3-5x\right|\)
=>B>=|5x-2+3-5x|=1
Dấu = xảy ra khi (5x-2)(5x-3)<=0
=>2/5<=x<=3/5
\(P=\sqrt[]{9x^2-6x+1}+\sqrt[]{25-30x+9x^2}\)
\(\Leftrightarrow P=\sqrt[]{\left(3x-1\right)^2}+\sqrt[]{\left(5-3x\right)^2}\)
\(\Leftrightarrow P=\left|3x-1\right|+\left|5-3x\right|\)
\(\Leftrightarrow P=\left|3x-1\right|+\left|5-3x\right|\ge\left|3x-1+5-3x\right|=4\)
Vậy \(GTNN\left(P\right)=4\)
\(A=x^2-4xy+7y^2+10x-24y+30\\ =\left(x^2-4xy+4y^2\right)+10\left(x-y\right)+25+\left(3y^2-14y+\dfrac{49}{3}\right)-\dfrac{34}{3}\\ =\left(x-2y+5\right)^2+3\left(y-\dfrac{7}{3}\right)^2-\dfrac{34}{5}\)
Với mọi x;y thì \(\left(x-2y+5\right)^2\ge0;3\left(y-\dfrac{7}{3}\right)^2\ge0\)
Do đó:\(A\ge-\dfrac{34}{5}\)
Để \(A=-\dfrac{34}{5}\) thì:
\(\left[{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-\dfrac{7}{3}\right)^2=0\\\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2y=-5\\y=\dfrac{7}{3}\\\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5+\dfrac{2.7}{3}=-\dfrac{1}{3}\\y=\dfrac{7}{3}\\\end{matrix}\right.\)
Vậy...
Nguyễn Thị Hồng Nhung, Akai Haruma, Trần Hoàng Nghĩa, Trần Thiên Kim, Phạm Hoàng Giang, Nhật Hạ, DƯƠNG PHAN KHÁNH DƯƠNG, Toshiro Kiyoshi, Ribi Nkok Ngok, ...