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a, 81 - y2 = 92 - y2 = ( 9 - y ).( 9 + y )
b, ( x - 3y )3 = x3 - 3x2.3y + 3x.( 3y )2 - ( 3y )3 = x3 - 9x2y + 27xy2 - 27y3
c, ( 2x + 2 )3 = ( 2x )3 + 3.( 2x )2.2 + 3.2x.22 + 23 = 8x3 + 24x2 + 24x + 8
a, 81 - y2 = 92 - y2 = ( 9 - y ).( 9 + y )
b, ( x - 3y )3 = x3 - 3x2 .3y + 3x.( 3y )2 - ( 3y )3 = x3 - 9x2y + 27xy2 - 27y3
c, ( 2x + 2 )3 = ( 2x )3 + 3.( 2x )2 .2 + 3.2x.22 + 23 = 8x3 + 24x2 + 24x + 8
ok thế là xong
\(a,\left(3x+1\right)^3=9x^3+9x^2+9x+1\)
\(b,\left(\frac{2}{3}x+1\right)^2=\frac{4}{9}x^2+\frac{4}{3}x+1\)
\(c,\left(x-y\right)^2-\left(x+y\right)^2=\left(x-y-x-y\right)\left(x-y+x+y\right)=-2y\cdot2x=-4xy\)
\(d,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y\cdot2x=4xy\)
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 :>
Phần a? phải là \(4a^2-4a+1\)chứ
a) \(4a^2-4a+1=\left(2a\right)^2+2.2a+1\)
\(=\left(2a+1\right)^2\)
b) \(9x^2-25y^2=\left(3x\right)^2-\left(5y\right)^2\)
\(=\left(3x-5y\right)\left(3x+5y\right)\)
c) \(1-2x+a^2=\left(1-a\right)^2\)
d) \(\left(2x+1\right)-2.\left(2x+1\right)\left(3x-y\right)+\left(3x-y\right)^2\)
\(=\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)
nếu có sai thì bn thông cảm
1.
b) nó là hằng đẳng thức rồi bn nhá
c) \(1-2a+a^2\)= \(1^2-2a1+a^2\)=\(\left(1-a\right)^2\)
d)\(\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)=\(\left(2x+1-3x+y\right)^2\)=\(\left(1-x+y\right)^2\)
2.
a)\(\left(\frac{1}{2}x\right)^2-\left(3y\right)^2\)=\(\left(\frac{x}{2}-3y\right)\left(\frac{x}{2}+3y\right)\)
b) Ko khai triển đc
c) \(4x^2+2xy+\frac{1}{4}y^2\)
a) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)
c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)
d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)
1, a, ( x2 - 1 )*( x2 + 2x)
= x4 + 2x3 - x2 - 2x
b, ( 2x - 1 )*( 3x + 2 )*( 3 - x )
= ( 6x2 + 4x - 3x - 2 )*( 3 - x )
= ( 6x2 + x - 2 )*( 3 - x )
= 18x2 - 6x3 + 3x - x2 - 6 + 2x
= -6x3 - 17x2 + 5x - 6
2, b, B = \(8x^3+48x^2+96x+64\)
= \(\left(2x+4\right)^3\)
Thay x = 8 vào B ta có:
B = \(\left(2\cdot8+4\right)^3\)
= ( 16 + 4 )3
= 203
= 8000
mình chỉ làm đc câu b thôi câu a bạn xem lại đề đi hình như câu a sai đề rồi
1,
\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)
2,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
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
a) ( x - 3 )( x2 + 3x + 9 )=x3-9
b) ( 5x - 1 ) ( 1 + 5x + 25x2 )=125x3-1
c) ( x2 - 1 ) ( x4 + x2 + 1 )=x6-1
a, \(\left(2x-1\right)^2=\left(2x\right)^2-2.2x.1+1^2=4x^2-4x+1\) 1
b, \(\left(3x+1\right)^2=\left(3x\right)^2+2.3x.1+1^2\) \(=9x^2+6x+1\)
c, \(\left(\frac{1}{2}x+1\right)^3\) = \(\left(\frac{1}{2}x\right)^3+3.\left(\frac{1}{2}x\right)^2.1+3.\frac{1}{2}x.1^2+1^3\)
= \(\frac{1}{8}x^3+\frac{3}{4}x^2+\frac{3}{2}x+1\)
d, \(-x^2+100\) = \(x^2+10^2\)