Tính giá trị biểu thức:

          a) x + (– 10), biết x = – 28

Với x = -28 thì x + (– 10) = -28 + (– 10) = -38

          b) (– 267) +  y, biết y = –  33

Với y = -33 thì (– 267) +  (-33) = - 300

a/ =-128

b/=(-300)

(-267) + (-33) = -(267 + 33) = -300

a) , biết

b) biết

(-267)+(-33)=(-300)

a) \(-28 + (-10 ) = -( 28 + 10 ) = -38\)

b) \(-267+\left(-33\right)=-\left(267+33\right)=-300\)

\(a,x+\left(-12\right)=\left(-24\right)+\left(-12\right)=-36\\ b,\left(-234\right)+y=\left(-234\right)+\left(-145\right)=-379\\ c,x+\left(-12\right)+\left(-234\right)=\left(-1\right)+\left(-12\right)+\left(-234\right)=-247\)

a) x=-12

b)y=89

Bài 5

a) A = -x³ + 6x² - 12x + 8

= -x³ + 3.(-x)².2 - 3.x.2² + 2³

= (-x + 2)³

= (2 - x)³

Thay x = -28 vào A ta được:

A = [2 - (-28)]³

= 30³

= 27000

b) B = 8x³ + 12x² + 6x + 1

= (2x)³ + 3.(2x)².1 + 3.2x.1² + 1³

= (2x + 1)³

Thay x = 1/2 vào B ta được:

B = (2.1/2 + 1)³

= 2³

= 8

Bài 6

a) 11³ - 1 = 11³ - 1³

= (11 - 1)(11² + 11.1 + 1²)

= 10.(121 + 11 + 1)

= 10.133

= 1330

b) Đặt B =  x³ - y³ = (x - y)(x² + xy + y²)

= (x - y)(x² - 2xy + y² + 3xy)

= (x - y)[(x - y)² + 3xy]

Thay x - y = 6 và xy = 9 vào B ta được:

B = 6.(6² + 3.9)

= 6.(36 + 27)

= 6.63

= 378

x+y-2=0

=>x+y=2

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

\(=x^2\left(x+y\right)-2x^2-y\left(x+y\right)+3y+x-5\)

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

=x+y-5

=2-5

=-3

a: \(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{x+y}{\sqrt{x}+\sqrt{y}}\)

Ta có: \(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{\left(x+y\right)_{}\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(\) \(=\frac{x^2-\sqrt{xy}\left(x+y\right)-y^2-x^2+y^2}{\sqrt{xy}\left(x-y\right)}=\frac{-\left(x+y\right)}{x-y}\)

b: Thay x=3; \(y=4+2\sqrt3\) vào A, ta được:

\(A=\frac{-\left(3+4+2\sqrt3\right)}{3-\left(4+2\sqrt3\right)}=\frac{-7-2\sqrt3}{-2\sqrt3-1}=\frac{7+2\sqrt3}{2\sqrt3+1}\)

\(=\frac{\left(7+2\sqrt3\right)\left(2\sqrt3-1\right)}{12-1}=\frac{14\sqrt3-7+12-2\sqrt3}{11}=\frac{12\sqrt3+5}{11}\)