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
tim gtln cua
a=-4x^2-8x+3
b=6x-x^2+2
c=x(2-3x)
d=3x-x^2+2
e=3-2x^2+2xy-y^2-2x
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
tim gtln cua
a=-4x^2-8x+3
b=6x-x^2+2
c=x(2-3x)
d=3x-x^2+2
e=3-2x^2+2xy-y^2-2x
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 = \(-\frac{3}{4}\)
tim gtln
1-x^2+4x
2000/x^2+2x+6
19-9x^2+6x
-x^2-4x-y^2+2y
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
d,5x+10/4x-8.4-2x/x+2
Bài 2: rút gọn
a, 6x ² y ³/8x ³y ²
b, x ³-x/3x+3
c, x ²+3xy/x ²-9y ²
d, x ²+4x+4/3x+6
Bài 3: Thực hiện phép tính
a, (x/x-3+(9-6x/x ²-3x)
b, 1/x-1/x+1
c, (x-12/6x-36)+(6/x ²-6x)
d, (6x-3/x):(4x ²-1/3x ²)
e, (x+y/2x-2y)-(x-y/2x+2y)-(y ²+x ²/y ²-x ²)
f, 7x+6/2x(x+7)-3x+6/2x ²+14x
g, (2/x+2-4/x ²+4x+4):(2/x ²-4+1/2-x)
Bài 1: Thực hiện phép tính
a) (x-4) (x+4) - (5-x) (x+1)
b) (3x^2 - 2xy + 4) + ( 5xy - 6x^2 - 7)
Bài 2: Rút gọn biểu thức
a) 3x^2 (2x + y) - 2y(4x^2 - y)
b) (x+3y) (x-2y) - (x^4 - 6x^2y^3): x^2y
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
Bài 2:
a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)
= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y2
= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y2
= 6\(x^3\) - 5\(x^2\)y + 2y2
tim x,y thuoc n de:
(x+22) chia het cho (x+1)
(2x+23) thuoc B *(x-1)
(3x+1) chia het cho (2x-1)
(x-2)*(2y+1)=17
xy+x+2y=5
( 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
a, x+22 chia hết cho x+1
suy ra : x+1+21 chia hêt cho x+1
mà x+1 chia hết cho x+1
suy ra 21 chia hết cho x+1
suy ra x+1 thuộc -1, 1 , 3, -3, 7, -7, 21, -21
suy ra x thuộc -2, 0, 2, -4, 6, -8, 20, -22
b, 2x+23 thuộc x-1
suy ra 2x+23 = x-1
2x-x= -23-1
x= -24
c, 3x+1 chia hết cho 2x-1
suy ra 2(3x+1) chia hết cho 2x-1
6x+2 chia hết cho 2x-1 (1)
lai có 2x-1 chia hết cho 2x-1
suy ra 3(2x-1) chia hết cho 2x-1
6x-3 chia hết cho 2x -1 (2)
từ 1 và 2
suy ra (6x+2)-(6x-3) chia hết cho 2x-1
5 chia hết cho 2x-1
suy ra 2x-1 thuộc -1,1,5,-5
x thuộc 0 , 1, 3, -2
c va d thì x thuộc z mới tìm được
K CHO MÌNH NHÉ
1. phan tich da thuc thanh nhan tu
a. x^2+3x-5 b. 4x^2-16x+7 c. 5x^2-6x-7 d.x^4+2x^3-4x-4
2. tim x,y bt: x^2+y^2+z^2=xy+yz+zx va x^2012+y^2012+z^2012= 3^2013
3. tim x: a. x^2-4x=21 b. x^2-4x+4=0 c.x^2-6x=2x=11 d. 4^x-12.2^x+32=0
1) Phân tích thành nhân tử:
a) x^4+2x^3-4x-4
b)x^2-2x-4y^2-4y
c)x^2(1-x^2)-4-4x^2
d)x^2+y^2-x^2y^2+xy-x-y
2) Phân tích thành nhân tử:
a)x^2+2x-24
b)x^2+3x+2
c)2x^2+3x+1
d)3x^2-4x+1
3) a) Tìm GTNN:
A=x^2+6x-5
B=x^2-3x+4
b) Tìm GTLN:
C= -x^2-2x+7
D= -3x^2-4x+2
\(x^2+3x+2\)
\(=x^2+x+2x+2\)
\(=x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
A = \(\dfrac{5xy^2-3z}{3xy}+\dfrac{4x^2y+3z}{3xy}\)
B = \(\dfrac{3y+5}{y-1}+\dfrac{-y^2-4y}{1-y}+\dfrac{y^2+y+7}{y-1}\)
C = \(\dfrac{6x}{x^2-9}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\)
D = \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
E = \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)