a: \(P=-5x^3+6x^2-2x\)

\(=-5\cdot\left(-1\right)^3+6\cdot\left(-1\right)^2-2\cdot\left(-1\right)\)

\(=-5\cdot\left(-1\right)+6+2=5+6+2=13\)

b: \(Q=-2\cdot\left(-\dfrac{1}{3}\right)^2\cdot\dfrac{11}{4}+4\cdot\dfrac{11}{4}+11\cdot\dfrac{1}{9}\cdot\dfrac{11}{4}\)

\(=-\dfrac{11}{2}\cdot\dfrac{1}{9}+11+\dfrac{121}{36}=\dfrac{55}{4}\)

a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

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

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

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)

\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)

\(=\dfrac{x-3}{3x}\)

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

a, Thay x = 1/2 ; y = -1/3 ta được 

\(A=\dfrac{3.1}{8}\left(-\dfrac{1}{3}\right)+\dfrac{6.1}{4}.\left(\dfrac{1}{9}\right)+\dfrac{3.1}{2}\left(-\dfrac{1}{3}\right)^3\)

\(=-\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{3}{2\left(-27\right)}=-\dfrac{7}{72}\)

b, Thay x = -1 ; y = 3 ta được 

\(B=9+\left(-1\right).3-1+27=32\)

bạn thay chỗ nào x là \(\dfrac{1}{2}\) còn chỗ nào y là \(\dfrac{-1}{3}\)nhé

còn như là 3\(x^3\)y thì thành là 3.\(x^3\).y nhé

mk lười nên ko giải ra cho bạn được 

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`

a: P(x)=6x^3-4x^2+4x-2

Q(x)=-5x^3-10x^2+6x+11

M(x)=x^3-14x^2+10x+9

b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)

=10x^4-11x^3-5x^2-15x+21

a: \(A=2\cdot2^2-\dfrac{1}{3}\cdot9=8-3=5\)

b: \(B=\dfrac{1}{2}a^2-3b^2=\dfrac{1}{2}\cdot4-3\cdot\dfrac{1}{9}=2-\dfrac{1}{3}=\dfrac{5}{3}\)

Thay x = 2 và y=9

A = 2.2² -\(\dfrac{1}{3}\).9

=  2.4 -\(\dfrac{1}{3}.9\)

= 8 - 3

= 5

Thay a = -2 và b = \(-\dfrac{1}{3}\)

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

B = \(\dfrac{1}{2}.4-3.\dfrac{1}{9}\)

B = \(2-\dfrac{1}{3}\)

B = \(\dfrac{5}{3}\)

B=12a2−3b2=12⋅4−3⋅19=2−13=53

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

Bài 6 :

a) \(11^3-1=\left(11-1\right)\left(11^2+11+1^2\right)\)

\(\)\(=10.\left(121+12\right)\)

\(=10.133\)

\(=1330\)

b) \(\left\{{}\begin{matrix}x-y=6\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2xy=36\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2.18=36\\xy=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=72\\xy=9\end{matrix}\right.\)

Ta có :

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

\(=6.\left(72+9\right)\)

\(=6.81\)

\(=486\)

x=4

=>x+1=5

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

  =x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1

  =x^6-x-1

  =4^6-4-1

  =4091

\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)

\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)

b)tương tự

=x^2006-x^2006-x^2005+x^2005+x^2004-...+x^3-x^2-x^2-x-5

=-x-5

=-7-5=-12

a: Khi x=-2 thì \(A=3\cdot\left(-2\right)^2+5\cdot\left(-2\right)-1=12-10-1=1\)

b: \(B=6xyz^4=6\cdot3\cdot2\cdot1^4=36\)

a) thay x = -2 và biểu thúc ta đc

\(3.\left(-2\right)^2+5.\left(-2\right)-1=12-10-1=1\)

b)thay x = 3 ; y=2 ,z=1 và biểu thúc ta đc

\(9.3.2.1^4+5.3.2.1^4-8.3.2.1^4=3.2.1^4\left(9+5-8\right)=6.6=36\)

a)=1
b)=36