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A = 2x2 + y2 - 2xy - 2y + 2000 = (x2 - 2xy + y2) + 2(x - y) + 1 + (x2 + 2x + 1) + 1998
= (x - y)2 + 2(x - y) + 1 + (x + 1)2 + 1998 = (x - y + 1)2 + (x + 1)2 1998 \(\ge\)1998 với mọi x,y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\x+1=0\end{cases}}\) <=> \(\hept{\begin{cases}y=x+1\\z=-1\end{cases}}\) <=> \(\hept{\begin{cases}x=-1\\y=0\end{cases}}\)
Vậy MinA = 1998 khi x = -1 và y = .0
b) B = x2 + 5y2 - 2xy + 6x - 18y + 50 = (x2 - 2xy + y2) + 6(x - y) + 9 + (4y2 - 12y + 9) + 32
= (x - y)2 + 6(x - y) + 9 + (2y - 3)2 + 32 = (x - y + 3)2 + (2y - 3)2 + 32 \(\ge\)32 với mọi x,y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y+3=0\\2y-3=0\end{cases}}\)<=> \(\hept{\begin{cases}x=y-3\\y=\frac{3}{2}\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)
Vậy MinB = 32 khi x = -3/2 và y = 3/2
c) C = 3x2 + x + 4 = 3(x2 + 1/3x + 1/36) + 47/12 = 3(x + 1/6)2 + 47/12 > = 47/12 với mọi x
Dấu "=" xảy ra <=> x + 1/6 = 0 <=> x = -1/6
Vậy MinC = 47/12 khi x = -1/6
A = 2y2 + x2 - 2xy - 2y + 2000 ( vầy mới tính được bạn nhé ;-; )
= ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + 1999
= ( x - y )2 + ( y - 1 )2 + 1999
\(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y\right)^2+\left(y-1\right)^2+1999\ge1999\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-1=0\end{cases}}\Leftrightarrow x=y=1\)
=> MinA = 1999 <=> x = y = 1
B = x2 + 5y2 - 2xy + 6x - 18y + 50
= ( x2 - 2xy + y2 + 2x - 6y + 9 ) + ( 4y2 - 12y + 9 ) + 32
= [ ( x2 - 2xy + y2 ) + 2( x - y ).3 + 32 ] + ( 2y - 3 )2 + 32
= [ ( x - y )2 + 2( x - y ).3 + 32 ] + ( 2y - 3 )2 + 32
= ( x - y + 3 ) + ( 2y - 3 )2 + 32
\(\hept{\begin{cases}\left(x-y+3\right)^2\ge0\forall x,y\\\left(2y-3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y+3\right)^2+\left(2y-3\right)^2+32\ge32\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y+3=0\\2y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)
=> MinB = 32 <=> x = -3/2 ; y = 3/2
C = 3x2 + x + 4
= 3( x2 + 1/3x + 1/36 ) + 47/12
= 3( x + 1/6 )2 + 47/12 ≥ 47/12 ∀ x
Đẳng thức xảy ra <=> x + 1/6 = 0 => x = -1/6
=> MinC = 47/12 <=> x = -1/6
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
bài 5 :
+) ta có : \(A=x^2-4x+18=x^2-4x+4+14\)
\(=\left(x-2\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(B=x^2-x+2=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\forall x\Rightarrow\left(đpcm\right)\)
+) ta có : \(C=x^2+2y^2-2xy-2y+15=x^2-2xy+y^2+y^2-2y+1+14\)
\(=\left(x-y\right)^2+\left(y-1\right)^2+14\ge14>0\forall x\Rightarrow\left(đpcm\right)\)
bài 6 :
+) ta có : \(M=x^2-10x+3=x^2-10x+25-22=\left(x-5\right)^2-22\ge-22\)
\(\Rightarrow M_{min}=-22\) khi \(x=5\)
+) ta có : \(N=x^2+6x-5=x^2+6x+9-14=\left(x+3\right)^2-14\ge-14\)
\(\Rightarrow N_{min}=-14\) khi \(x=-3\)
+) ta có : \(P=x^2+y^2-4x+20=x^2-4x+4+y^2+16=\left(x-2\right)^2+y^2+16\ge16\)
\(\Rightarrow P_{min}=16\) khi \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
+) ta có : \(Q=x\left(x-3\right)=x^2-3x=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)
\(\Rightarrow Q_{min}=\dfrac{-9}{4}\) khi \(x=\dfrac{3}{2}\)
bài 7 :
+) ta có : \(A=-x^2-12x+3=-\left(x^2+12x+36\right)+39=-\left(x+6\right)^2+39\le39\)
\(\Rightarrow A_{max}=39\) khi \(x=-6\)
+) ta có : \(B=-4x^2+4x+7=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\le11\)
\(\Rightarrow B_{max}=11\) khi \(x=2\)
bài 8 :
a) ta có : \(16x^2-9=0\Leftrightarrow x^2=\dfrac{9}{16}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
b) ta có : \(\left(x-2\right)^2-x^2=4\Leftrightarrow x^2-4x+4-x^2-4=0\Leftrightarrow x=0\)
c) ta có : \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=\dfrac{-255}{2}\)
d) ta có : \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)=16\)
\(\Leftrightarrow4x^2-12x+9-4x^2+1-16=0\Leftrightarrow-12x-6=0\Leftrightarrow x=\dfrac{-1}{2}\)
e) ta có : \(\left(x-2\right)\left(x+2\right)-x\left(x-2\right)=1\)
\(\Leftrightarrow x^2-4-x^2+2x-1=0\Leftrightarrow x=\dfrac{5}{2}\)
a: \(2x^3+x^2-13x+6\)
\(=2x^3-4x^2+5x^2-10x-3x+6\)
\(=\left(x-2\right)\left(2x^2+5x-3\right)\)
\(=\left(x-2\right)\left(2x^2+6x-x-3\right)\)
\(=\left(x-2\right)\left(x+3\right)\left(2x-1\right)\)
b: \(2x^2+y^2-6x+2xy-2y+5=0\)
\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)
=>x-2=0 và x+y-1=0
=>x=2 và y=-1
a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
\(2D=x^2-4xy+4y^2+x^2-12x+36+6y^2-36y+54+10\)\(2D=\left(x-2y\right)^2+\left(x-6\right)^2+6\left(y-3\right)^2+10\)
\(2D\ge10\) => D>=5 khi x=2y=6
\(F=3x^2+x+4=3\left(x^2+\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{47}{12}\)
F=\(3\left(x+\dfrac{1}{6}\right)^2+\dfrac{47}{12}\ge\dfrac{47}{12}\) khi x=-1/6
\(2E=4x^2-4xy+y^2+y^2-4y+4+3996\)
\(2E=\left(2x-y\right)^2+\left(y-2\right)^2+3996\ge3996\)
E>=1998 khi 2x=y=2
bài 4;
\(B=-3x^2+x=-3\left(x^2-\dfrac{2x}{6}+\dfrac{1}{36}\right)+\dfrac{1}{12}\)
\(B=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}\le\dfrac{1}{12}\)
khi x=1/6
bài 5:
\(a,\left(x+2\right)^2=0=>x=-2\)
\(b,\left(x-6\right)^2+\left(y+1\right)^2=0\rightarrow\left\{{}\begin{matrix}x=6\\y=-1\end{matrix}\right.\)
c,\(x^2+2y^2-2xy-2x+2=0\)
\(x^2-4xy+4y^2+x^2-4x+4=0\)
\(\left(x-2y\right)^2+\left(x-2\right)^2=0\rightarrow\left\{{}\begin{matrix}x=2y\\x=2\end{matrix}\right.\)
đây nhá bạn, khá tốn time của mình