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\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
a, (452 - 2.40.45 + 402) - 152
= ( 45 - 40 )2 - 152
= 52 - 152 = ( 5 - 15 )( 5 + 15 )
= -200
b, 13 . 4 . 13 .11 - 13 . 4 . 13 . 3 - 32
= 132 . 44 - 132 . 12 - 32
= 132 ( 44 -12 ) - 32
= 32 ( 132 - 1 )
= 32 . ( 13 - 1 )( 13 + 1 )
= 32 . 12 . 14
= 5376
\(45^2+40^2-15^2-80\cdot45\)
\(=\left(45^2-2\cdot45\cdot40+40^2\right)-15^2\)
\(=\left(45-40\right)^2-15^2\)
\(=15^2-15^2\)
\(=0\)
\(52\cdot143 -52\cdot39-8\cdot4\)
\(=7436-2028-32\)
\(=5408-32\)
\(=5440\)
a)\(x^2+7x+6\)
\(=x^2+6x+x+6\)
\(=x\left(x+6\right)+\left(x+6\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
b)\(x^4+2016x^2+2015x+2016\)
\(=x^4+2016x^2+\left(2016x-x\right)+2016\)
\(=\left(x^4-x\right)+\left(2016x^2+2016x+2016\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
Bài 3:
Từ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Rightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\) (1)
Ta thấy:\(\begin{cases}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{cases}\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (2)
Từ (1) và (2) \(\Rightarrow\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\)
\(\Rightarrow\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=1\\c=1\end{cases}\)
\(\Rightarrow a=b=c=1\Rightarrow H=1\cdot1\cdot1+1^{2014}+1^{2015}+1^{2016}=1+1+1+1=4\)
\(\left(a^2-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a+1\right)\left(a^2-a+1\right)\)
\(=\left(a^3-1\right)\left(a^3+1\right)=a^6-1\)
1, \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right)\left(4x+3\right)\)
2,\(x^2-4+\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)^3\)
3,\(5a\left(a-2\right)-a+2=5a\left(a-2\right)-1\left(a-2\right)=\left(5a-1\right)\left(a-2\right)\)
4,\(7\left(a-5\right)+8a\left(5-a\right)=7\left(a-5\right)-8a\left(a-5\right)=\left(7-8a\right)\left(a-5\right)\)
5, \(25a^2-4b^2+4b-1=25a^2-\left(4b^2-4b+1\right)=\left(5a\right)^2-\left(2b-1\right)^2=\left(5a-2b+1\right)\left(5a+2b-1\right)\)
Ta có: \(x+y=2\Rightarrow x^2+2xy+y^2=4\Rightarrow x^2+y^2=4-2xy\)
Mặt khác: \(\frac{\left(x+y\right)^2}{4}\ge xy\)
\(\Rightarrow1\ge xy\) (thay x+y=2) và \(2\ge2xy\)
Ta có: \(xy\left(x^2+y^2\right)=xy\left(4-2xy\right)\)
=>.....
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
\(1^2-2^2+3^2-4^2+...+999^2-1000^2\)
\(=\left(1-2\right)\left(1+2\right)+\left(1-2\right)\left(3+4\right)+...+\left(999-1000\right)\left(999+1000\right)\)
\(=-\left(1+2+3+4+...+999+1000\right)\)
\(=-\dfrac{\left(\dfrac{1000-1}{1}+1\right).\left(1000+1\right)}{2}=-500500\)
Không làm :)