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
Bài 2 :
a, Ta có :
\(\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
=> đpcm
b, Ta có :
\(\left(a+b\right)^2-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\)
=> đpcm
c, Ta có :
\(\left(ax-by\right)^2+\left(ay+bx\right)^2\)
\(=a^2x^2-2abxy+b^2y^2+a^2y^2+2abxy+b^2x^2\)
\(=\left(a^2x^2+b^2x^2\right)+\left(-2abxy+2abxy\right)+\left(b^2y^2+a^2y^2\right)\)
\(=\left(a^2+b^2\right)x^2+\left(b^2+a^2\right)y^2\)
\(=\left(a^2+b^2\right)\left(x^2+y^2\right)\)
=> đpcm
Học tốt nhé
Bài 1.
a) A = x2 - y2 = ( x - y )( x + y ) = ( 17 - 13 )( 17 + 13 ) = 4.30 = 120
b) B = ( 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
Bài 2
a) VP = a2 - 2ab + b2 + 4ab = a2 + 2ab + b2 = ( a + b )2 = VT
b) VP = a2 + 2ab + b2 - 4ab = a2 - 2ab + b2 = ( a - b )2 = VT
c) VT = a2x2 + a2y2 + b2x2 + b2y2 = ( a2x2 - 2axby + b2y2 ) + ( a2y2 + 2axby + b2x2 ) = ( ax - by )2 + ( ay + bx )2 = VP
