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26 tháng 8 2017
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) \(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)
PT
26 tháng 8 2017
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,\(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)
DB
8 tháng 9 2020
a, (x+y)2 = x2 + 2xy + y2
b, ( x-4y)2= x2 -8xy2 + 16y2
c, \(\left(3x+\frac{1}{3}\right)^2=9x^2+2xy+\frac{1}{9}\)
d,\(4x^2-81=\left(2x-9\right)\left(2x+9\right)\)
e,\(\left(xy+5\right)^2=x^2y^2+10xy+25\)
f,\(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)
g,\(1-9y^2=\left(1-3y\right)\left(1+3y\right)\)
h,\(\left(m-\frac{2}{3}n\right)^2=m^2-\frac{4}{3}mn+\frac{4}{9}n^2\)
LT
19 tháng 9 2020
Bài 1: Khai triển các hằng đẳng thức
a) ( x - 3 )( x2 + 3x + 9 )
= x3 - 33
= x3 - 27
b) ( 5x - 1 )( 1 + 5x + 25x2 )
= ( 5x - 1 )(25x2 + 5x + 1 )
= (5x)3 - 1
= 125x3 - 1
c) ( x2 - 1 ) ( x4 + x2 + 1 )
= (x2)3 - 1
= x6 - 1
OF
19 tháng 9 2020
a) ( x - 3 )( x2 + 3x + 9 )=x3-9
b) ( 5x - 1 ) ( 1 + 5x + 25x2 )=125x3-1
c) ( x2 - 1 ) ( x4 + x2 + 1 )=x6-1
GP
0
TG
15 tháng 8 2020
a) \(4x^2-12x+9\)
\(=\left(2x\right)^2-2.2x.3+3^2\)
\(=\left(2x-3\right)^2\)
b) \(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
c) \(1+12x+36x^2\)
\(=1^2+2.1.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2\)
\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)
\(=\left(3x-4y\right)^2\)
e) \(\frac{x^2}{4}+2xy+4y^2\)
\(=\left(\frac{x}{2}\right)^2+2.\frac{x}{2}.2y+\left(2y\right)^2\)
\(=\left(\frac{x}{2}+2y\right)^2\)
f) \(-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.5x+5^2\right)\)
\(=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\left[\left(4b\right)^2+2.4b.3a+\left(3a\right)^2\right]\)
\(=-a^4b^4\left(4b+3a\right)^2\)
h) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2.5x^2y+y^2\)
\(=\left(5x^2-y\right)^2\)
16 tháng 9 2021
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
NH
1
4 tháng 9 2020
Áp dụng công thức : (A + B)3 = A3 + 3A2B + 3AB2 + B3
(A - B)3 = A3 - 3A2B + 3AB2 -B3
a) (3x + 1)3 = (3x)3 + 3.(3x)2.1 + 3.3x.1 + 13 = 27x3 + 27x2 + 9x + 1
b) \(\left(\frac{x}{3}-1\right)^3=\left(\frac{x}{3}\right)^3-3\cdot\left(\frac{x}{3}\right)^2\cdot1+3\cdot\left(\frac{x}{3}\right)\cdot1^2-1^3\)
\(=\frac{x^3}{27}-3\cdot\frac{x^2}{9}\cdot1+3\cdot\frac{x}{3}\cdot1-1\)
= \(\frac{x^3}{27}-\frac{x^2}{3}+x-1\)
c) \(\left(2x-\frac{1}{x}\right)^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot\frac{1}{x}+3\cdot2x\cdot\left(\frac{1}{x}\right)^2-\left(\frac{1}{x}\right)^3\)
\(=8x^3-3\cdot4x^2\cdot\frac{1}{x}+6x\cdot\frac{1}{x^2}-\frac{1}{x^3}\)
\(=8x^3-12x+\frac{6}{x}-\frac{1}{x^3}\)
d) \(\left(-y^2+3x\right)^3=\left(3x-y^2\right)^3=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot y^2+3\cdot3x\cdot y^4-y^6\)
= 27x3 - 27x2y2 + 9xy4 - y6
= -y6 + 9xy4 - 27x2y2 + 27x3
Tương tự câu cuối :>
ND
1
6 tháng 8 2023
\(A=\left(x^2-4y^2\right)\left(x^2-2xy+4y^2\right)\left(x^2+2xy+4y^2\right)\)
\(A=\left(x-2y\right)\left(x+2y\right)\left(x^2-2xy+4y^2\right)\left(x^2+2xy+4y^2\right)\)
\(A=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(A=\left[x^3-\left(2y\right)^3\right]\left[x^3+\left(2y\right)^3\right]\)
\(A=\left[x^3-8y^3\right]\left[x^3+8y^3\right]\)
\(A=x^6-64y^6\)