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a) \(=\left(x-2\right)^2\)
b) \(=\left(3x-2\right)^2\)
c) \(=\left(x-3y\right)^2\)
d) \(=\left(\dfrac{x}{2}+1\right)^2\)
e) \(=\left(x-4\right)^2\)
f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)
g) \(=\left(x-1\right)\left(x+1\right)\)
h) \(=\left(5x-4\right)\left(5x+4\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)\left(\dfrac{3}{4}+\dfrac{1}{2}x\right)\)
\(a,a^2\left(a-b\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ c,=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ b,=\left(x-5\right)^2-9y^2=\left(x-5-3y\right)\left(x-5+3y\right)\\ d,=4\left(x^2-9x+14\right)=4\left(x-7\right)\left(x-2\right)\)
a: \(\left(3x-1\right)\left(9x^2+3x+1\right)=27x^3-1\)
b: \(\left(1-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{5}+1\right)=1-\dfrac{x^3}{125}\)
c: \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
d: \(\left(4x+3y\right)\left(16x^2-12xy+9y^2\right)=64x^3+27y^3\)
- Đức và Thọ đều viết đúng;
Hương nhận xét sai;
- Sơn rút ra được hằng đẳng thức là: (x - 5)2 = (5 - x)2
`9x^2+4y^2-12xy+6x-4y+1`
`=(3x)^2-2.3x.2y+(2y)^2+2(3x-2y)+1`
`=(3x-2y)^2+2(3x-2y)+1`
`=(3x-2y+1)^2`
a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
a)\(-25+4x^2=\left(2x-5\right)\left(2x+5\right)\)
b)\(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
c)\(\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)
\(a,-25+4x^2=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(b,-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
\(c,\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x\right)^2+\frac{2.1}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)(sửa đề)