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Bài 1:
\(VT=1\cdot\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\cdot\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^4-b^4\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=a^{32}-b^{32}\)
1) \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
2) \(\left(2x-3\right)^2=4x^2-12x+9\)
\(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(2x-5\right)^2=4x^2-20x+25\)
\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)
3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)
\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)
\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}\)
\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)
1: \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
(x^2+2x+1)-(y^2+4y+4)=6
(x+1)^2-(y+2)^2=6
(x+1-y-2)(x+1+y+2)=6
(x-y-1)(x+y+3)=6
nhân ra làm tiếp đc ko e
Bài 4:
Ta có: \(\left(4n+3\right)^2-25\)
\(=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)\)
\(=8\left(n+2\right)\left(2n-1\right)⋮8\)
Bài 4:
Ta có: ( 4n + 3 )2 - 25
= ( 4n + 3 - 5 ) . ( 4n + 3 + 5 )
= ( 4n - 2 ) . ( 4n + 8 )
= 8 ( n + 2 ) . ( 2n - 1 ) dấu chia hết 8
Mik ko viết đc dấu chia hết nhé
a. (4-x+4)(4+x-4)
=(8-x)x
b, (x+y)^2-2z(x+y)
=(x+y)(x+y-2z)
hok tốt
nha
lỗi