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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,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
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) \(4x^2+2xy+\frac{1}{4}y^2\)
\(x+3^2=x^2+2.x.3+3^2=x^2+6x+9\)
\(4x^{^{ }2}-9=4x^2-2.4x.9+9^2=16x^2-72x+81\)
\(\left(x-\dfrac{3}{2}\right)^2=x^2-2.x.\dfrac{3}{2}+\dfrac{3}{2}^2=x^2-3x+\dfrac{9}{4}\)
\(\left(x+3\right)^2=x^2+6x+9\)
\(\left(4x^2-9\right)^2=16x^4-72x^2+81\)
\(\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
\(\left(x-\dfrac{3}{2}\right)^2=x^2-3x+\dfrac{9}{4}\)
\(x^2-4=\left(x+2\right)\left(x-2\right)\)
\(x^2-289=\left(x+17\right)\left(x-17\right)\)
1.
a)\(\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
b)\(9a^2+3ab+\frac{1}{4}a^2\)
2.
a)\(\left(5x+2b\right)^2\)
b)\(\left(x+1\right)^2\)
c)\(\left(3x+1\right)^2\)
d)\(\left[\left(2x+3y\right)+1\right]^2\)
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 :>
câu 1:
\(a,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
=> \(25x^2+10x+1-\left(25x^2-9\right)=30\)
=> \(25x^2+10x+1-25x^2+9=30\)
=> \(10x+10=30\)
=> \(10x=20\)
=> \(x=2\)
Vậy..........
\(b,\left(2x+3\right)^2-\left(2x-3\right)^2+4\left(x^2-6x\right)=64\)
=> \(6.4x+4x^2-24x=64\)
=> \(24x+4x^2-24x=64\)
=> \(4x^2=64\)
=> \(x^2=64:4=16\)
=> \(\left|x\right|=\sqrt{16}\)
=> \(x=\pm4\)
Vậy \(x\in\left\{4;-4\right\}\)
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)
x3+3x2+3x+1
b)8x3+18x2+54x+27
c)x3+3636x2+3636x+1780360128
d)x6-6x2+12x-8
e)
8x3-36x2y+54xy2-27y3