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Chứng minh rằng
a/ (a+b)^2=(a-b)^2+4ab
b/ (a-b)^2=(a+b)^2-4ab
c/ (a^2+b^2)(x^2+y^2)=(ax-by)^2+(ay+bx)^2
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)
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 :
\(\left(2-1\right)B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow\left(2-1\right)B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow1.B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow B=2^{32}-1\)
a) (a+b)2 = (a-b)2 +4ab
⇔ (a+b)2 = a2 - 2ab + b2 +4ab
⇔ (a+b)2 = a2 + 2ab + b2
⇔ (a+b)2 = (a+b)2
⇒ (a+b)2 = (a-b)2 +4ab (dpcm)
b) (a-b)2 = (a+b)2 - 4ab
⇔ (a-b)2 = a2 + 2ab + b2 - 4ab
⇔ (a-b)2 = a2 - 2ab + b2
⇔ (a-b)2 = (a-b)2
⇒ (a-b)2 = (a+b)2 - 4ab (dpcm)