a: A=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+98+...+2+1

=5050

b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1

\(=2^{64}-1+1=2^{64}\)

BÀi 1:

a: \(\left(a+b\right)^2-\left(a-b\right)^2\)

=(a+b-a+b)(a+b+a-b)

\(=2b\cdot2a=4ab\)

b: \(\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)

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

\(=a^2+4a+4-a^2+4=4a+8\)

c: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

=>\(4x^2+12x+9-4\left(x^2-1\right)=49\)

=>\(4x^2+12x+9-4x^2+4=49\)

=>12x+13=49

=>12x=36

=>x=3

d: \(Q=\left(x+3\right)^2+\left(x+3\right)\left(x-3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)

\(=2x^2+6x-2\left(x^2-2x-8\right)=2x^2+6x-2x^2+4x+16=10x+16\)

Khi x=1/2 thì Q=10*1/2+16=5+16=21

Bài 2:

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

\(=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)

b: \(\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=49x^2+14x+1-\left(x^2+14x+49\right)\)

\(=49x^2+14x+1-x^2-14x-49=48x^2-48=48\left(x^2-1\right)\)

c: \(16x^2-\left(4x-5\right)^2=15\)

=>\(16x^2-\left(16x^2-40x+25\right)=15\)

=>\(16x^2-16x^2+40x-25=15\)

=>40x=40

=>x=1

d: \(A=-x^2+2x+3\)

\(=-x^2+2x-1+4\)

\(=-\left(x-1\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)

\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

Ta có

(a+b+c)²+(b+c-a)²+(c+a-b)²+(a+b-c)²= [(a+b)+c]²+[(b-a)+c]²+[(a-b)+c]²+[(a+b)-c]

=(a+b)²+2c(a+b)+c²+(b-a)²+2c(b-a)+c²+(a-b)²+2c(a-b)+c²+(a+b)²-2c(a+b)+c²

=2(a+b)²+2(a-b)²+4c²( vì (a-b)²=(b-a)²)

xàm quá bạn ơi

a) (a+b)³+(a-b)³=a³+3a²b+3ab²+b³+a³-3a²b+3ab²-b³

                          =2a³+6ab²

b) (a + b + c)² + (a − b − c)² + (b − c − a)² + (c − a − b)²

=a²+b²+c²+2ab+2bc+2ca+a²+b²+c²-2ab+2bc-2ac+a²+b²+c²-2bc+2ca-2ba+a²+b²+c²-2ca+2ab-2cb

=4a²+4b²+4c²

a) Ta có: \(\left(a+b\right)^3+\left(a-b\right)^3\)

\(=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2a\cdot\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)

\(=2a\cdot\left(a^2+3b^2\right)\)

\(=2a^3+6ab^2\)

hình như bạn chép sai đề