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1) a^2 + b^2 + 2a - 2b - 2ab = (a^2 - 2ab + b^2) + (2a-2b) = (a-b)^2 + 2(a-b) = (a-b)(a-b+2)
2) 4a^2 - 4b^2 - 4a + 1 = ( 4a^2 - 4a +1) - 4b^2 = (2a-1)^2 - 4b^2 = (2a-1-2b)(2a-1+2b)
3) a^3+6a^2+12a+8= (a^3+8)+(6a^2+12a)= (a+2)(a^2-2a+4)+6a(a+2)=(a+2)(a^2-2a+4+6a)=(a+2)(a^2+4a+4)=(a+2)(a+2)^2=(a+2)^3
\(b,=1^2-\left(x-y\right)^2=\left(1+x-y\right)\left(1-x+y\right)\)
\(c,=\left(x^2+1\right)^2-\left(2x\right)^2=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\)
1-x-2x^2
= 1-x-2x.2x
= 1 - ( x + 2x.2x)
= 1 - 5x
Để 1-x-2x^2 mang giá trị lớn nhất thì x phài là số âm.
\(A=1-x-2x^2\)
\(=-2\left(x^2+2\times x\times\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2-\frac{1}{2}\right)\)
\(=-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)
\(\left(x+\frac{1}{4}\right)^2\ge0\)
\(\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\ge-\frac{9}{16}\)
\(-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\le\frac{9}{8}\)
Vậy Max A = \(\frac{9}{8}\) khi x = \(-\frac{1}{4}\)
a: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b: \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
\(x^6-2x^3y+y^2=\left(x^3-y\right)^2\)
b: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
\(-a^2-2a-1=-\left(a+1\right)^2\)
Bài 1:
\(\left\{{}\begin{matrix}a=5c+1\\b=5d+2\end{matrix}\right.\)
\(a^2+b^2=\left(5c+1\right)^2+\left(5d+2\right)^2\)
\(=25c^2+10c+1+25d^2+20d+4\)
\(=25c^2+25d^2+10c+20d+5\)
\(=5\left(5c^2+5d^2+2c+4d+1\right)⋮5\)
Bài 3:
a: \(4x^2+12x+15=4x^2+12x+9+6=\left(2x+3\right)^2+6>=6\forall x\)
Dấu '=' xảy ra khi x=-3/2
b: \(9x^2-6x+5=9x^2-6x+1+4=\left(3x-1\right)^2+4>=4\forall x\)
Dấu '=' xảy ra khi x=1/3
a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)