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Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
a) \(\left(2x-3y\right)^2=4x^2-12xy+9y^2\)
b) \(\left(5p-q\right)^2=25p^2-10pq+q^2\)
c) \(\left(-a-b\right)^2=-a^2-2ab-b^2\)
d) \(\left(1+3s\right)^2=1+6s+9s^2\)
e) \(\left(a^2b+2b\right)^2=a^4b^2+4a^2b^2+4b^2\)
f) \(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)
a,\(\left(2x-3y\right)=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\)
=\(4x^2-12xy+6y^2\)
b,\(\left(5p-q\right)^2=\left(5p\right)^2-2.5p.q+q^2\)
=\(25p^2-10pq+q^2\)
c,(-a-b)\(^2=\left(-a\right)^2-2.\left(-a\right).b+b^2\)
=\(a^2+2ab+b^2\)
d,\(\left(1+3s\right)^2=1+6s+9s^2\)
e,(a\(^2b+2b)^2=(a^2b)^2+2.a^2b.2b^2+\left(2b\right)^2\)
=\(a^4b^2+4a^2b^2+4b^2\)
f,\(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)
1, Ta có:
(x+1)(x+2)(x+3)(x+4) - 15 = [(x+1)(x+4)].[(x+2)(x+3)] - 15
= (x^2+5x+4)(x^2+5x+6) - 15
Đặt x^2+5x+5 = y
=> (y-1)(y+1) - 15 = y^2 - 1 -15 = y^2 - 16 = (y-4)(y+4)
= (x^2+5x+1)(x^2+5x+9)
2,
Ta có : 2(a+b)(a-b) + (a-b)^2 + (a+b)^2 - 4b^2
= [(a+b)^2+2(a+b)(a-b)+(a-b)^2] - (2b)^2
= (a+b+a-b)^2 - (2b)^2 = (2a)^2 - (2b)^2 = (2a-2b)(2a+2b)
A=a(a+b)-b(a+b)=(a+b)(a-b)=a2-b2(hằng đẳng thức 3)
B=(3x+2)2+(3x-2)2-2(9x2-4)+x
=(3x+2)2+(3x-2)2-2(3x-2)(3x+2)+x
=[(3x+2)-(3x-2)]2+x
=42+x
=16+x
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
1. \(A=\left(a+b\right)^2+\left(a+b\right)^2\)
\(\Leftrightarrow A=2\left(a^2+2ab+b^2\right)\)
\(\Leftrightarrow A=2a^2+4ab+2b^2\)
2. \(B=\left(a+b\right)^2-\left(a-b\right)^2\)
\(\Leftrightarrow B=\left(a+b+a-b\right)\left(a+b-a+b\right)\)
\(\Leftrightarrow B=2a.2b=4ab\)
1) \(A=\left(a+b\right)^2+\left(a+b\right)^2\)
\(A=a^2+2ab+b^2+a^2+2ab+b^2\)
\(A=2a^2+4ab+2b^2\)
2) \(B=\left(a+b\right)^2-\left(a-b\right)^2\)
\(B=a^2+2ab+b^2-a^2+2ab-b^2\)
\(B=4ab\)