Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

Bài 2:

\(x+y+1=0\Rightarrow x+y=-1\)

A = \(x\)(\(x+y\)) - y².(\(x+y\)) + \(x^2\) - y² + 2(\(x+y\)) + 3

Thay \(x\) + y  = -1 vào biểu thức A ta có:

A = \(x\).( -1) - y² .(-1) + \(x^2\)  - y² + 2(-1) + 3

A = -\(x\) + y² + \(x^2\) - y² - 2 + 3

A = \(x^2\) - \(x\) + 1

a) Các biểu thức: \(\dfrac{1}{5}x{y^2}{z^3}; - \dfrac{3}{2}{x^4}{\rm{yx}}{{\rm{z}}^2}\) là đơn thức

b) Các biểu thức: \(2 - x + y; - 5{{\rm{x}}^2}y{z^3} + \dfrac{1}{3}x{y^2}z + x + 1\) là đa thức

bài 1:

|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1

a

+) A = 2x\(^2\) - 3x + 5

= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5

= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)

+) A = 2x\(^2\) - 3x + 5

= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5

= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5

= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)

b) +) B = 2x\(^2\) - 3xy + y\(^2\)

= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)

= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1

= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)

+) B = 2x\(^2\) - 3xy + y\(^2\)

= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)

= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1

= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)

bài 3

x.y.z = 2 và x + y + z = 0

A = ( x + y )( y +z )( z + x )

= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )

= 0 + 2 = 2

bài 4

a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)

=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)

+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)

=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)

x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)

+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)

2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0

x = 0 : 2 = 2

1 a

2c

3b

4d

5c

6c

9/4...... p ko????

BÀI 2 a, x²+x+1=(x²+1/2*2*x+1/4)-1/4+1=(x+1/2)² +3/4

MÀ (x+1/2)²>=0 với mọi giá trị của x .Dấu"=" xảy ra khi x+1/2=0 =>x=-1/2

    =>(x+1/2)²+3/4>=3/4 với mọi giá trị của x .Dấu "=" xảy ra khi x=-1/2

   =>x²+x+1 có giá trị nhỏ nhất là 3/4 khi x=-1/2

   b,A=y(y+1)(y+2)(y+3)

=>A =[y(y+3)] [(y+1)(y+2)]

  =>A=(y²+3y) (y²+3y+2)

Đặt X=y²+3y+1

=>A=(X+1)(X-1)

=>A=X²-1

=>A=(y²+3y+1)²-1

MÀ (y²+3y+1)²>=0 với mọi giá trị của y

=>(y²+3y+1)²-1>=-1

Vậy GTNN của Alà -1

c,B=x³+y³+z³-3xyz

=>B=(x³+y³)+z³-3xyz

=>B=(x+y)³-3xy(x+y)+z³-3xyz

=>B=[(x+y)³+z³]-3xy(x+y+z)

=>B=(x+y+z)(x²+2xy+y²-xz-yz+z²)-3xy(x+y+z)

=>B=(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)

=>B=(x+y+z)(x²+y²+z²-xy-xz-yz)

rút gọn các đa thức bằng cách nhân chúng với nhau rồi thay số vào là tính được mà

a) A = (x - 3)(x + 7) - (2x - 5)(x - 1)

       = x(x + 7) - 3(x + 7) - 2x(x - 1) + 5(x - 1)

      = x² + 7x - 3x - 21 - 2x² + 2x + 5x - 5

      = (x² - 2x²) + (7x - 3x + 2x + 5x) + (-21 - 5) = -x² + 11x - 26 = -(x² - 11x + 26)

+) Với x = 0 thì -(0² - 11.0 + 26) = -(0 - 0 + 26) = -26

+) Với x = 1 thì -(1² - 11.1 + 26) = -(1 - 11 + 26) = -16

b) B = (3x + 5)(2x - 1) + (4x - 1)(3x + 2)

       = 3x(2x - 1) + 5(2x - 1) + 4x(3x + 2) - 1(3x + 2)

       = 6x² - 3x + 10x - 5 + 12x² + 8x - 3x - 2

       = (6x² + 12x²) + (-3x + 10x + 8x - 3x)+ (-5 - 2) = 18x² + 12x - 7

|x| = 2 => x = 2 hoặc x = -2

Với x = 2 thì 18.2² + 12.2 - 7 = 18.4 + 24 - 7 = 72 + 24 - 7 = 89

Với x = -2 thì 18.(-2)² + 12.(-2) - 7 = 18.4 + (-24) - 7 = 18.4 - 24 - 7 = 41

c) C = (2x + y)(2z + y) + (x - y)(y - z)

= 2x(2z + y) + y(2z + y) + x(y - z) - y(y- z)

= 4xz + 2xy + 2zy + y² + xy - xz - y² + yz

= 4xz + 2xy + 2zy + (y² - y²)  +xy - xz + yz

= 4xz + 3xy + 3zy 

Với x = 1,y = 1,z = 1

= 4.1.1 + 3.1.1 + 3.1.1 = 4 + 3 + 3 = 10