1) PTTNT
a) x^2 - 4x^2y + 4xy
b)x^2 + 3x + x - 3y
2) Tim GTLN
-2x^2 + 3x - 5
3) tim x,y thuoc z
3xy + 6x - y = 7
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a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)
Vậy Max(A) = 7 khi x = -1
b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy Max(B) = 11 khi x = 3
c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{1}{3}\right)^2=0\Rightarrow x=\frac{1}{3}\)
Vậy Max(C) = 1/3 khi x = 1/3
d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy Max(D) = 17/4 khi x = 3/2
e) \(E=3-2x^2+2xy-y^2-2x\)
\(E=-\left(x^2-2xy+y^2\right)-\left(x^2+2x+1\right)+4\)
\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)
Vậy Max(E) = 4 khi x = y = -1
A = \(4x^2\) - 8x + 3
= [\(\left(2x\right)^2\) - 2.2x.2 + \(2^2\)] \(-2^2\) + 3
= \(\left(2x-2\right)^2\) - 1
Ta có: \(\left(2x-2\right)^2\) ≤ 0 ∀ x
\(\left(2x-2\right)^2\) - 1 ≤ - 1
Hay A ≤ - 1
Dấu "=" xảy ra ↔ 2x - 2 = 0
2x = 2
x = 1
Vậy GTLN của A = - 1 ↔ x = 1
B = 6x \(-x^2\) + 2
= - (\(x^2\) - 6x) + 2
= - (\(x^2\) - 2.x.3 + \(3^2\)) \(-3^2\) + 2
= - \(\left(x-3\right)^2\) -7
Ta có: \(-\left(x-3\right)^2\) ≤ 0 ∀ x
\(-\left(x-3\right)^2\) - 7 ≤ - 7
Hay B ≤ - 7
Dấu "=" xảy ra ↔ - (x - 3) = 0
- x + 3 = 0
- x= - 3
x = 3
Vậy GTLN của B = - 7 ↔ x = 3
C = x(2 - 3x)
= 2x \(-3x^2\)
= - 3(\(x^2\) - \(\frac{3}{2}x\) )
= - 3(\(x^2\) - 2.x.\(\frac{3}{4}\) + \(\frac{3}{4}^2\)) \(-\frac{3}{4}^2\)
Ta có: \(-3\left(x+\frac{3}{4}\right)^2\) ≤ 0 ∀ x
\(-3\left(x+\frac{3}{4}\right)^2\) \(-\frac{9}{16}\) ≤ \(-\frac{9}{16}\)
Hay C ≤ \(-\frac{9}{16}\)
Dấu "=" xảy ra ↔ \(-3\left(x+\frac{3}{4}\right)\) = 0
- 3x \(-\frac{9}{4}\) = 0
- 3x = \(\frac{9}{4}\)
x = \(-\frac{3}{4}\)
Vậy GTLN của C = \(-\frac{9}{16}\) ↔ x = \(-\frac{3}{4}\)
a: \(-x^2+4x+1\)
\(=-\left(x^2-4x-1\right)\)
\(=-\left(x^2-4x+4-5\right)\)
\(=-\left(x-2\right)^2+5\le5\)
Dấu '=' xảy ra khi x=2
b: \(x^2+2x+6=\left(x+1\right)^2+5\)
\(\Leftrightarrow\dfrac{2000}{\left(x+1\right)^2+5}\le400\)
Dấu '=' xảy ra khi x=-1
c: \(-9x^2+6x+19\)
\(=-\left(9x^2-6x-19\right)\)
\(=-\left(9x^2-6x+1-20\right)\)
\(=-\left(3x-1\right)^2+20\le20\)
Dấu '=' xảy ra khi x=1/3
d: \(=-\left(x^2+4x+y^2-2y\right)\)
\(=-\left(x^2+4x+4+y^2-2y+1-5\right)\)
\(=-\left(x+2\right)^2-\left(y-1\right)^2+5\le5\)
Dấu '=' xảy ra khi x=-2 và y=1
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
( x + 22 ) \(⋮\)( x + 1 )
x + 1 + 21 \(⋮\)( x + 1 )
Mà x + 1 \(⋮\)x + 1 → 21 \(⋮\)x + 1 \(\in\)Ư ( 21 )
( x - 2 ) . ( 2y + 1 ) = 17
Mà 17 là số nguyên tố và bằng 1 . 17
→ Nếu ( x - 2 ) = 1 thì ( 2y + 1 ) = 17
→ Nếu ( 2y + 1 ) = 1 thì ( x - 2 ) = 17
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Bài 2:
\(A=-2x^2+3x-5\)
\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)
\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)
Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)
Bài 1:
a)x2-4x2y+4xy
=x(x-4xy+y)
b)đề sai