a) \(\left(a+b\right)^2=a^2+2ab+b^2\left(1\right)\)

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

Từ (1) và (2) => đpcm

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

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

Từ (3) và (4) =>đpcm

c) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(5\right)\)

\(\left(ax-by\right)^2+\left(ay+bx\right)^2=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(6\right)\)

Từ (5) và (6) =>đpcm

a) VP=(a-b)²+4ab

        =a²-2ab+b²+4ab

        =a²+b²+2ab

        =(a+b)²=VT

Vậy (a+b)^2 = (a-b)^2 +4ab

b) VP=(a+b)²-4ab

        =a²+2ab+b²-4ab

        =a²-2ab+b²

        =(a-b)²=VT

Vậy (a-b)^2 = (a+b)^2 - 4ab

c)

VP=(ax-by)²+(ay+bx)²

=a²x²-2axby+b²y²+a²y²+2axby+b²x²

=a²x²+b²y²+a²y²+b²x²

=(a²x²+b²x²)+(b²y²+a²y²)

=x².(a²+b²)+y².(a²+b²)

=(a²+b²)(a²+y²)=VT

Vậy ( a^2 + b^2 ).(x^2 +y^2) = (ax - by)^2 +(ay+bx)^2

a) Ta có: \(\left(a+b\right)^2=4ab\)<=> \(a^2+b^2+2ab=4ab\)

                                               <=> \(a^2-2ab+b^2=0\)

                                                <=> \(\left(a-b\right)^2=0\)=> a=b (đpcm)

b) Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

<=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

<=> \(a^2y^2+b^2x^2-2axby=0\)

<=>\(\left(ay-bx\right)^2=0\)

<=>ay=bx(đpcm)

b)(a-b)^2
=a^2 -2ab+b^2
=a^2 +2ab+b^2 -4ab
=(a+b)^2 - 4ab
a)(a+b)^2
=a^2 +2ab+b^2
=a^2 -2ab+b^2 +4ab
=(a-b)^2 + 4ab

c)a^3+b^3

=(a^3+3a^2b+3ab^2+b^2)-(3a^2b+3ab^2)

=(a+b)^3-3ab(a+b)

d)a^3-b^3

=(a^3-3a^2b+3ab^2-b^3)+(3a^2b-3ab^2)

=(a-b)^3+3ab(a-b)

e)(a^2+b^2)(x^2+y^2)

=(a.x)^2+(b.x)^2+(a.y)^2+(b.y)^2

=((a.x)^2-2abxy+(b.y)^2)+((a.y)^2-2abxy+(b.x)^2)

=(ax-by)^2+(ay+bx)^2

l-ike giùm mik vs công sức cả buổi đấy

a) Sửa đề: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

Ta có: \(VP=\left(a-b\right)^2+4ab\)

\(=a^2-2ab+b^2+4ab\)

\(=a^2+2ab+b^2\)

\(=\left(a+b\right)^2=VT\)(đpcm)

b) Ta có: \(VT=\left(a-b\right)^2\)

\(=a^2-2ab+b^2\)

\(=a^2+2ab+b^2-4ab\)

\(=\left(a+b\right)^2-4ab=VP\)(đpcm)

c) Ta có: \(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2\)

\(=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)

\(=a^2x^2+b^2y^2+a^2y^2+b^2x^2\)

\(=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)

\(=\left(x^2+y^2\right)\left(a^2+b^2\right)=VT\)(đpcm)

f

1. \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

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

\(\Rightarrow VT=VP\)

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

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

\(\Rightarrow VT=VP\)

3. \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(bx+ay\right)^2\)

\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(VP=\left(ax-by\right)^2+\left(bx+ay\right)^2=a^2x^2-2axby+b^2y^2+b^2x^2+2bxay+a^2y^2=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\Rightarrow VT=VP\)

Câu 1:

A=x^2- y^2=(x-y)(x+y)

Thay x=17, y=13 vào A, ta có: A= (17-13)(17+13)=4.30=120

=> Vậy A=120 tại x=17,y=13.

b, B= (2+1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1) (đề bài đúng)

      = 1.(2+1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1) 

      = (2-1)(2+1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1) 

      = (2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1) 

      = (2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1) 

      = (2⁸-1)(2⁸+1)(2¹⁶+1) 

       = (2¹⁶-1) (2¹⁶+1)

       = 2³²-1

=> B= = 2³²-1

Bài 1 :

a,Ta có :

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

\(=\left(x-y\right)\left(x+y\right)\)

Với x = 17 và y = 13 ta có :

\(A=\left(17-13\right)\left(17+13\right)\)

\(=4.30\)

\(=120\)

Vậy x = 120 với x = 17 và y = 13 .

b, Nhân biểu thức đã cho với ( 2 - 1 ) ta được :

\(\left(2-1\right)B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow\left(2-1\right)B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow1.B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow B=2^{32}-1\)

             Bài làm :

Bài 1 :

a) A=x² - y² =(x-y)(x+y)=(17-13)(17+13)=4.30=120

b) Sửa đề tí nhéa

 \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(B=\left(2^{32}-1\right)\)

Bài 2 :

\(a\text{)}VT=\left(a+b\right)^2=a^2+2ab+b^2=\left(a^2-2ab+b^2\right)+4ab=\left(a-b\right)^2+4ab=VP\)

=> Điều phải chứng minh

\(b\text{)}VT=\left(a-b\right)^2=a^2-2ab+b^2=\left(a^2+2ab+b^2\right)-4ab=\left(a+b\right)^2-4ab=VP\)

=> Điều phải chứng minh

c) Ta có :

\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2=a^2x^2+b^2y^2+a^2y^2+b^2x^2\)

=> VT = VP

=> Điều phải chứng minh