V Junior High School math problem induction

Finish a piece of algebra

P1

Calculate $\left (\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\right) \div \left (\ Frac{1}{1007}+\frac{1}{1008}+...+frac{1}{2012}\right) $

The dividing sequence of the problem is subtracted from the dislocation.

Set
S1 = 1+1/3+1/5+1/7+1/9+...+1/2011
S2 = -1/2-1/4-1/6-1/8-...-1/2012
   = 1/2+1/4+1/6+...+1/2012-(1+1/2+1/3+1/4+...+1/1006)
So set
SA = 1/2+1/4+...+1/2012
SB = 1+1/2+1/3+...+1/1006
Then the order of the semicolon above is listed as
S1 + SA-SB
S1 + SA = 1+1/2+1/3+1/4+...+1/2012
And then minus the SB.
S1 + SA-SB = 1/1007+1/1008+...+1/2012
The original answer to the score mother is 1

P2

Calculation

\[\sqrt[3]{\frac{\sqrt{5}-1}{2}+\left (\frac{\sqrt{5}-1}{2}\right) ^2} \]

The problem is quite simple.

Observation (sqrt (5)-1)/2 is the AC this, the square is the ad this, each other into a golden separation, then their difference DC is ac^3, opened three square root is just AC.

P3 for any $x,y,z\in \mathbb{r}$, define operations ' $\otimes$ ' as

\[x \otimes y = \frac{3x^3y+3x^2y^2+xy^3+45} {(x+1) ^3+ (y+1) ^3-60}\]

This operation is combined from left to right, that is, $ (x\otimes y\otimes z= (x\otimes y) \otimes z) $

Please

\[2013\otimes 2012\otimes 2011\otimes ... \otimes 3\otimes 2\]

This question is more conscience orz.

Still offering the observation of Dafa.

Note that for any $x\otimes 3$, there are $x\otimes 3=9$

Then just ask for the $9\otimes 2=\frac{5463}{967}$.

P4 known

\[X-y=6,\sqrt{x^2-xy}+\sqrt{xy-y^2}=9\]

The value of the $\sqrt{x^2-xy}-\sqrt{xy-y^2}$.

This silly question, a second glance.

Looking at the relationship between known and unknown, we can obviously apply the squared difference formula.

\[\left (\sqrt{x^2-xy}+\sqrt{xy-y^2}=9\right) \left (\sqrt{x^2-xy}-\sqrt{xy-y^2}\right) = (x-y) ^2 = 36 \]

Then the original formula is $36\div 9=4$.

P5 set $abc=1$, seek the value of $\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}$.

This problem is very interesting.

may wish to set
A = 1/x
So
BC = X
So
B=x/c
Brought into the original type.
Primitive $=\frac {\frac {1} {x}} {\frac {1} {c} + \frac{1}{x}+1}+\frac{\frac{x}{c}}{x+\frac{x}{c}+1}+\frac{c}{\frac{c}{x }+x+1}$
On the molecular-pass, it can be multiplied.
Primitive $=\frac {c} {c+x+1} +\frac {x} {c+x+1} +\frac {1} {c+x+1} $
It's a shenmegui?!.
is not the legendary $1$ of the famous.

P6 known $x,y,z\in \mathbb{r},\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1$, beg $\frac {x^2} {y+z} +\frac {y^2} {x+z} +\frac {z^2} The value of {x+y} =1$.

The problem is very, very interesting.

The standard solution of this problem is to mess with.

We-pass the first formula, the two sides of the equation multiply on the molecule, messing with a way to eliminate the same items on both ends, surprised to find ...

\[-xyz=x^3+y^3+z^3\]

Fun, yes! (interested readers can see for themselves)

The-pass of the desired formula is basically the same as the above.

The problem is this-pass, do not do, how to do!!!

We'll tidy it up.

(The following steps write only the denominator)

$x ^{4}\quad +\quad x^{3}y\quad +\quad x^{2}yz\quad +\quad xy^{3}\quad +\quad y^{4}\quad +\quad y^{3}z\quad +\q Uad xy^{2}z\quad +\quad z^{2}xy\quad +\quad z^{4}\quad +\quad z^{3}y\quad +\quad xz^{3}$

=

$x ^4+y^4+z^4+x^3y+x^3z+xy^3+y^3z+z^3y+z^3x+\left (x+y+z\right) xyz$

=

$x ^4+y^4+z^4+x^3y+x^3z+xy^3+y^3z+z^3y+z^3x+\left (x+y+z\right) \left (-x^3-y^3-z^3\right) $

=

$x ^4+y^4+z^4+x^3y+x^3z+xy^3+y^3z+z^3y+z^3x-\left (x^4+y^4+z^4+x^3y+x^3z+xy^3+y^3z+z^3y+z^3x\right) $

=

$0$

Isn't it funny?

Nimon feel at random ...

P7

This problem is really very interesting.

Nest eruption to $\left (x^2-x-2\right) ^6$ is a 12-time polynomial. We remember it as

\[a_{12}x^{12}+a_{11}x^{11}+a_{10}x^{10}+...+a_1x+a_0\]

So what we're asking for IS

\[a_{12}+a_{10}+a_8+a_6+a_4+a_2\]

I thought I was going to have an aldehyde-based interpolation.

And look at the back of this series 233.

I was still yy. A polynomial-power algorithm that does not require a fast exponentiation 23333.

We write the function as $f\left (x\right) $.

Remember

\[\begin{array}{lll}s_1 = f (1) & & \mathtt{(1)}\s_2 = f (0) & & \mathtt{(2)}\s_3 = f (-1) & & \mathtt{(3)}\end{array}\]

So,

$f (1) =a_{12}+a_{11}+a_{10}+...+a_1+a_0=64$

$f (0) =a_0=64$

$f ( -1) =a_{}-a_{}+a_{-...-a_{1}+a_{0}=0$

Hand play out the original =

$\frac{f (1) -2f (0) +f (-1)}{2}=-32$

Very interesting ^ ^ (hey that polynomial, your lap is really messy)

P8

Decomposition-dependent

\[x^2-2x-2y^2+4y-xy\]

V Junior High School math problem induction