B=5x2+4xy-2(x-2y)+2y2+3
=5x2+4xy-2x+4y+2y2+3
=(4x2+4xy+y2)+(x2-2x+1)+(y2+4y+4)-2
=(2x+y)2+(x-1)2+(y+2)2-2 \(\ge\) -2
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+y=0\\x-1=0\\y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
a) Đặt A = u2 + v2 - 2u + 3v + 15
= (u2 - 2u + 1) + (v2 + 3v + 9/4) + 47/4
= (u - 1)2 + (v + 3/2)2 + 47/4 \(\ge\frac{47}{4}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}u-1=0\\v+\frac{3}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}u=1\\v=-\frac{3}{2}\end{cases}}\)
Vậy Min A = 47/4 <=> u = 1 ; y = -3/2
\(P=2x^2+5y^2+4xy+8x-4y+15\)
\(=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-5\)
\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-5\)
Ta có :
\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) \(\Leftrightarrow P\ge-5\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy \(P_{Min}=-5\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
\(D=2023-8x+2y+4xy-y^2-5x^2\)
\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)
\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)
\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)
\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)
\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)
Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)
\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
Đặt A = 5x2 + 2y2 + 4xy - 2x + 4y + 2022
= (2x2 + 4xy + 2y2) + 4(x + y) + 2 + (3x2 - 6x + 3) + 2017
= 2(x + y)2 + 4(x + y) + 2 + 3(x - 1)2 + 2017
= 2(x + y + 1)2 + 3(x - 1)2 + 2017 \(\ge\)2017
=> Min A = 2017
\(5x^2+2y^2+4xy-2x+4y+2022\)
\(=\left(4x^2+4x+y^2\right)+\left(y^2+4y+4\right)+\left(x^2-2x+1\right)+2017\)
\(=\left(2x+y\right)^2+\left(y+2\right)^2+\left(x-1\right)^2+2017\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+y=0\\y+2=0\\x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Vậy \(Min_A=2017\Leftrightarrow x=1;y=-2\)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
2:
a: \(3xy^2-3x^3-6xy+3x\)
\(=3x\cdot\left(y^2-2y+1-x^2\right)\)
\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)
=3x(y-1-x)(y-1+x)
b: \(3x^2+11x+6\)
\(=3x^2+9x+2x+6\)
=3x(x+3)+2(x+3)
=(x+3)(3x+2)
c: \(-x^3-4xy^2+4x^2y+16x\)
\(=x\left(16+4xy-4y^2-x^2\right)\)
\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)
=x(4-x+2y)(4+x-2y)
d: \(xz-x^2-yz+2xy-y^2\)
=z(x-y)-\(\left(x^2-2xy+y^2\right)\)
=\(z\left(x-y\right)-\left(x-y\right)^2\)
=(x-y)(z-x+y)
e: \(4x^2-y^2-6x+3y\)
=(2x-y)(2x+y)-3(2x-y)
=(2x-y)(2x+y-3)
f: \(x^4-x^3-10x^2+2x+4\)
\(=x^4+2x^3-2x^2-3x^3-6x^2+6x-2x^2-4x+4\)
\(=\left(x^2+2x-2\right)\left(x^2-3x-2\right)\)
g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)
\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)
\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)
\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)
=(x-1)(x^2+1)(11-5y-1)(11+5y+1)
=(x-1)(x^2+1)(10-5y)(12+5y)
=5(2-y)(x-1)(x^2+1)(5y+12)
Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
Mấy cái kia tương tự
\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)
A=5x2+2y2−4xy−8x−4y+19=(2x2−4xy+2y2)+4(x−y)+(3x2−12x)+19=2(x−y)2+4(x−y)+3(x2−4x+4)+7=2[(x−y)2+2(x−y)+1]+3(x−2)
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mik viết 5x2 là 5x mũ 2 nha