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bài này nhìn cái là ra nghay mà phân tích 26=25+1 bỏ 25 vào x bình với 10x , bỏ 1 vào y bình và 2y là ok
a ) Ta có : -x3 + 3x2 - 3x + 1
= 1 - 3x + 3x2 - x3
= (1 - x)3
b) Ta có : 8 - 12x + 6x2 - x3
= 23 - 3.22.x + 3.2.x2 - x3
= (2 - x)3
a, -x3 + 3x2 - 3x + 1
= -x3 + 3.x2.1 - 3.x.12 + 13
= ( -x + 1 )3
a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
\(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)
\(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b\right)^2\)
a)Chú ý đề em sai nha!
\(x^2-16xy+64y^2\)
\(=x^2-2.x.8y+\left(8y\right)^2\)
\(=\left(x-8y\right)^2\)
b) \(16x^2y^2+40xy+25\)
\(=\left(4xy\right)^2+2.4xy.5+5^2\)
\(=\left(4xy+5\right)^2\)
a) \(x^2-16xy-64y^2\)
\(=x^2-16xy+64y^2-128y^2\)
\(=\left(8y-x\right)^2-\left(\sqrt{128}x\right)^2\)
\(=\left(8y-x-\sqrt{128}x\right)\left(8y-x+\sqrt{128}x\right)\)
b) \(16x^2y^2+40xy+25\)
\(=\left(4xy\right)^2+2.4xy.5+5^2\)
\(=\left(4xy+5\right)^2\)
a) x2 - 4x + 2 = (x2 - 4x + 4) - 2 = (x - 2)2 - 2 = \(\left(x-2+\sqrt{2}\right)\left(x-2-\sqrt{2}\right)\)
b) x2 - 12x + 11 = x2 - x - 11x + 11 = x(x - 1) - 11(x - 1) = (x - 1)(x - 11)
c) 3x2 + 6x - 9 = 3x2 - 3x + 9x - 9 = 3x(x - 1) + 9(x - 1) = (3x + 9)(x - 1) = 3(x + 3)(x - 1)
d) 2x2 - 6x + 2 = 2(x2 - 3x + 1) = 2(x2 - 3x + 9/4 - 5/4) = 2[(x - 3/2)2 - 5/4] = \(2\left(x-\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\left(x-\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\)
1.
a) \(x^2-4x+2=\left(x^2-4x+4\right)-2=\left(x-2\right)^2-2=\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)\)
b) \(x^2-12x+11=\left(x^2-12x+36\right)-25=\left(x-6\right)^2-5^2=\left(x-6-5\right)\left(x-6+5\right)=\left(x-11\right)\left(x-1\right)\)
c) \(3x^2+6x-9=3\left(x^2+2x-3\right)=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2-6x+2=2\left(x^2-3x+1\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]\)
\(=2\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)
a) (x+3)2
b) (x-8)2
a) x2 + 6x + 9
= x2 + 3x + 3x + 9
= x(x + 3) + 3(x + 3)
= (x + 3)(x + 3) = (x + 3)2
b) x2 + 16x + 64
= x2 + 8x + 8x + 64
= x(x + 8) + 8(x + 8)
= (x + 8)(x + 8) = (x + 8)2