\(a,A=y^2-\dfrac{1}{2}y+\dfrac{1}{16}\)

\(=y^2-2.y.\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\)

\(=\left(y-\dfrac{1}{4}\right)^2\)

Với \(y=100,25\), ta được:

\(A=\left(100,25-\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{401}{4}-\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{400}{4}\right)^2=100^2=10000\)

\(------\)

\(b,B=4x^2-9y^2-6y-1\)

\(=\left(2x\right)^2-\left[\left(3y\right)^2+2.3y.1+1\right]\)

\(=\left(2x\right)^2-\left(3y+1\right)^2\)

\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)

Với \(x=23;y=1\), ta được:

\(B=\left(2.23-3.1-1\right)\left(2.23+3.1+1\right)\)

\(=\left(46-4\right)\left(46+4\right)\)

\(=42.50=2100\)

a: A=5x^2y-5x^2y-3xy+2xy+xy+x^4y^2+1+x^2

=x^4y^2+x^2+1

Khi x=-1 và y=1 thì A=(-1)^4*1^2+(-1)^2+1=3

b: A=x^2(x^2y^2+1)+1>=1>0 với mọi x,y

=>A luôn dương với mọi x,y

1.

a) = 354

b) = -54

c) = -167

2.

A= -37

a. Thay a = 14, b = 15, c = 10, ta có:

\(a=a\times b+200\)

\(=>a=14\times15+200\)

\(=>a=210+200=410\)

___

\(b=a\times b\times c\)

\(=>b=14\times15\times10=2100\)

b. Vì 410 < 2100 nên a < b.

\(#NqHahh\)

a: Khi a=14 và b=15 thì \(A=14\cdot15+200=210+200=410\)

Khi a=14 và b=15 và c=10 thì \(B=14\cdot15\cdot10=210\cdot10=2100\)

b: A=410

B=2100

=>A<B

Câu 1. C

Câu 2. B

Câu 3. D

Câu 4. B

Câu 5. B

A=x^2+6x+9+1

=(x+3)^2+1

Thay x=-103 vào A, ta được:

A=(-103+3)^2+1=10000+1=10001

\(a,A=x^2+6x+10\)

\(=\left(x^2+2\cdot x\cdot3+3^2\right)+1\)

\(=\left(x+3\right)^2+1\)

Thay \(x=-103\) vào \(A\), ta được:

\(A=\left(-103+3\right)^2+1\)

\(=\left(-100\right)^2+1\)

\(=10000+1\)

\(=10001\)

#Urushi

Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

\(=x^2-3x+6x-12\)

\(=x^2+3x-12\)

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

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

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

d: \(x\left(xy-1\right)\left(xy+1\right)\)

\(=x\left(x^2y^2-1\right)\)

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

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

Bài 3: 

Ta có: \(A=\left(x+2\right)\left(3x-1\right)-x\left(3x+3\right)-2x+7\)

\(=3x^2-x+6x-2-3x^2-9x-2x+7\)

=5

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349