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\(a)\)
\(4x^2-4x+1\)\(=\left(2x-1\right)^2\)
\(b)\)
\(\left(3x+2\right)\left(2-3x\right)\)\(=4-9x^2\)
\(c)\)
\(\left(x-3\right)\left(x^2+3x+9\right)\)\(=x^3-27\)
\(a,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1=\left(2x-1\right)^2\)
\(b,\left(3x+2\right)\left(2-3x\right)=\left(2+3x\right)\left(2-3x\right)=2^2-\left(3x\right)^2\)
\(c,\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+3x.1+3^2\right)=x^3-3^3\)
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)
a. x^2 + 5x + 25/4 = x^2 + 2.x .5/2 + (5/2)^2 = ( x + 5/2)^2
b, 16x^2 - 8x + 1 = (4x)^2 - 2.4x.1 + 1 = ( 4x - 1 )^2
c, 4x^2 + 12xy + 9y^2 = (2x)^2 + 2.2x.3y + (3y)^2 = ( 2x + 3y)^2
Bài giải
\(a,\text{ }a^2+9-6a=a^2+2\cdot3a+3^2=\left(a-3\right)^2\)
\(b,\text{ }x^2-x+\frac{1}{4}=x^2-2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
\(c,\text{ }-x^2+4x-x=3x-x^2=\left(\sqrt{3x}\right)^2-x^2=\left(\sqrt{3x}-x\right)\left(\sqrt{3x}+x\right)\)( Đề nói vận dụng hằng đẳng thức để rút gọn nên mình đưa về hiệu hai ình phương nha ! )
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) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-x^4-16x+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-y^4+4=y^4-81-y^2+4=-77\)
a)=x^2-4
b)=(x+2)^2
c)=(2x-1)^2
d)=x^2-5^2=(x-5)(x+5)
a, =x2 - 4
b, ( x + 2)2
c, ( 2x - 1) 2
d, = (x-5)(x+5)