2016 APE Tutoring Junior High School Math contest training Camp homework answer-7

Scan the following two-dimensional code to download and install the Ape Tutorial app, please search the teacher's name "Zhao Yin" after opening to enroll in This course (14 lessons, 99 yuan).

1. If $3x^2-x = 1$, what is the value of $6x^3 + 7x^2-5x + 2016$?

Answer: $$6x^3 + 7x^2-5x + = 2x (3x^2-x-1) + 9x^2-3x + 2016$$ $$= 9x^2-3x + = 3 (3x^2-x -1) + 2019 = 2019.$$

2. What is the minimum value of the polynomial $2x^2-4xy + 5y^2-12y + 13$?

Answer: $$2x^2-4xy + 5y^2-12y + 2 (x^2-2xy + y^2) + 3 (y^2-4y + 4) $$ $$= 2 (x-y) ^2 + 3 (y-2) ^2 + 1 \ge 1.$ The equals sign when and only if $x = y = 2$.

Therefore, the minimum value of the polynomial is $1$.

3. If $x + y + z = a$, $xy + yz + ZX = b$, $xyz = C $, with $a, B, C $ for $xy ^2 + x^2y + yz^2 + y^2z + z^2x + x^2z$.

Answer: $ $a \cdot B = (x^2y + xyz + x^2z) + (xy^2 + y^2z + xyz) + (xyz + yz^2 + xz^2) $$ $$= xy^2 + x^2y + yz^2 + y^2z + z^2x + x^2z + 3xyz$$ $$\rightarrow xy^2 + x^2y + yz^2 + y^2z + z^2x + x^2z = ab-3c.$$

4. Equation $ (x^2 + 3x-4) ^2 + (2x^2-7x + 6) ^2 = (3x^2-4x + 2) What is the solution of ^2$?

Answer:

Note that $ (x^2 + 3x-4) + (2x^2-7x + 6) = 3x^2-4x + 2$, so consider swapping elements.

Make $x ^2 + 3x-4 = m$, $2x^2-7x + 6 = n$, then $ $m ^2 + n^2 = (m + N) ^2 \rightarrow mn = 0.$$ This can be $ $x ^2 + 3x-4 = 0\rightar Row (x+4) (x-1) = 0 \rightarrow x_1 = -4,\ x_2 = 1,$$ or $$2x^2-7x + 6 = 0 \rightarrow (2x-3) (x-2) = 0 \rightarrow x _3 = {3\over2},\ X_4 = 2.$$ fully, the solution of the original equation is: $ $x _1 = -4,\ x_2 = 1,\ x_3 = {3\over2},\ X_4 = 2.$$

5. Set $x + y = 1$, $x ^2 + y^2 = 2$, $x ^7 + y^7$ value.

Answer: $ $xy = {1\over2}\left[(x+y) ^2-(x^2 + y^2) \right] =-{1\over2}$$ $$\rightarrow x^3 + y^3 = (x+y) (x^ 2-xy + y^2) = {5\over2},$$ $ $x ^4 + y^4 = (x^2 + y^2) ^2-2x^2y^2 = {7\over2}$$ $$\rightarrow x^7 + y^7 = (x^3 + y^3) (x^4 + y^4)-x^3y^3 (x + y) = {71\over 8}.$$

6. Known $m, n$ are natural numbers and $m \ne n$. Verification: Natural number $m ^4 + 4n^4$ must be expressed as the sum of the squares of four natural numbers.

Answer: $ $m ^4 + 4n^4 = (m^2 + 2n^2) ^2-4m^2n^2$$ $$= (m^2 + 2n^2 + 2mn) (m^2 + 2n^2-2mn) $$ $$= \left[(m+n) ^2 + n^2\right]\left[(m-n) ^2 + n^2\right]$$ $$= (m+n) ^2 (m-n) ^2 + n^2 (m+n) ^2 + (m-n) ^2n^2 + n^4$$ $$= (m^2-n^2) ^2 + (MN + n^2) ^2 + (mn-n^2) ^2 + (n^2) ^2.$$

2016 APE Tutoring Junior High School Math contest training Camp homework answer-7