Đề thi giữa kì 1 Toán 8 Kết nối tri thức – Đề 2 (có đáp án)

Ta có $A+B-C=\left(5x^2-2xy+3x{y}^2\right)+\left(-2x^2-2x{y}^2+xy\right)-\left(x^2-3x^{2}y+xy-2x^3\right)$

                  $=5x^2-2xy+3x{y}^2-2x^2-2x{y}^2+xy-x^2+3x^{2}y-xy+2x^3$

                  $=2x^3+\left(5x^2-2x^2-x^2\right)+\left(-2xy+xy-xy\right)+\left(3x{y}^2-2x{y}^2\right)+3x^{2}y$

                  $=2x^3+2x^2-2xy+x{y}^2+3x^{2}y$.

$A-2B+C=\left(5x^2-2xy+3x{y}^2\right)-2\left(-2x^2-2x{y}^2+xy\right)+\left(x^2-3x^{2}y+xy-2x^3\right)$

                  $=5x^2-2xy+3x{y}^2+4x^2+4x{y}^2-2xy+x^2-3x^{2}y+xy-2x^3$

                  $=-2x^3+\left(5x^2+4x^2+x^2\right)+\left(-2xy-2xy+xy\right)+\left(3x{y}^2+4x{y}^2\right)-3x^{2}y$

                  $=-2x^3+10x^2-3xy+7x{y}^2-3x^{2}y$.

a) Ta có $$35^2-700+10^2=35^2-\mathrm{2.35.10}+10^2={\left(35-10\right)}^2=25^2=625$$.

b) Ta có $$\frac{{12}^3+8^3}{{15}^3+5^3}$$$=\frac{\left(12+8\right).\left({12}^2-12.8+8^2\right)}{\left(15+5\right).\left({15}^2-15.5+5^2\right)}$$=\frac{20.\left(144-96+64\right)}{20.\left(225-75+25\right)}$$=\frac{112}{175}=\frac{16}{25}$.

Ta có $x^3-3x^{2}y+3x{y}^2-y^3={\left(x-y\right)}^3$.

Thay $x=77$ và $y=-23$ vào biểu thức trên ta được ${\left(77-\left(-23\right)\right)}^3=100^3=1000000$.

Vậy giá trị của biểu thức đã cho tại $x=77$ và $y=-23$ là $1000000$.

Ta có

$$A=a^2\left(a^2+b^2\right)\left(a^2+c^2\right)$$

$=\left(a^4+a^2{b}^2\right)\left(a^2+c^2\right)$

$=a^6+a^4{c}^2+a^4{b}^2+a^2{b}^2{c}^2$

$=a^4\left(a^2+c^2+b^2\right)+a^2{b}^2{c}^2$

$=a^4.0+a^2{b}^2{c}^2$

$=a^2{b}^2{c}^2$.           (1)

$B=b^2\left(b^2+c^2\right)\left(b^2+a^2\right)$

$=\left(b^4+b^2{c}^2\right)\left(b^2+a^2\right)$

$=b^6+b^4{a}^2+b^4{c}^2+a^2{b}^2{c}^2$

$=b^4\left(b^2+a^2+c^2\right)+a^2{b}^2{c}^2$

$=b^4.0+a^2{b}^2{c}^2$

$=a^2{b}^2{c}^2$.           (2)

$C=c^2\left(c^2+a^2\right)\left(c^2+b^2\right)$

$=\left(c^4+c^2{a}^2\right)\left(c^2+b^2\right)$

$=c^6+c^4{b}^2+c^4{a}^2+a^2{b}^2{c}^2$

$=c^4\left(c^2+b^2+a^2\right)+a^2{b}^2{c}^2$

$=c^4.0+a^2{b}^2{c}^2$

$=a^2{b}^2{c}^2$.           (3)

Từ (1), (2) và (3) suy ra $A=B=C$.

$a^2+b^2-a^2{b}^2+ab-a-b$

$=\left(a^2-a^2{b}^2\right)-\left(b-b^2\right)-\left(a-ab\right)$

$=a^2\left(1-b^2\right)-b\left(1-b\right)-a\left(1-b\right)$

$=a^2\left(1-b\right)\left(1+b\right)-b\left(1-b\right)-a\left(1-b\right)$

$=\left(1-b\right)\left(a^2+a^{2}b-b-a\right)$

$=\left(1-b\right)\left[\left(a^2-a\right)+\left(a^{2}b-b\right)\right]$

$=\left(1-b\right)\left[a\left(a-1\right)+b\left(a-1\right)\left(a+1\right)\right]$

$=\left(1-b\right)\left(a-1\right)\left(a+ab+b\right)$.

$xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)$

$=x^{2}y+x{y}^2-y^{2}z-y{z}^2+xz\left(x-z\right)$

$=\left(x^{2}y-y{z}^2\right)+\left(x{y}^2-y^{2}z\right)+xz\left(x-z\right)$

$=y\left(x^2-z^2\right)+y^2\left(x-z\right)+xz\left(x-z\right)$

$=y\left(x-z\right)\left(x+z\right)+y^2\left(x-z\right)+xz\left(x-z\right)$

$=\left(x-z\right)\left(xy+yz+y^2+xz\right)$

$=\left(x-z\right)\left[\left(xy+xz\right)+\left(yz+y^2\right)\right]$

$=\left(x-z\right)\left[x\left(y+z\right)+y\left(y+z\right)\right]$

$=\left(x-z\right)\left(y+z\right)\left(x+y\right)$.

Ta có $$a=\underset{n}{\underbrace{\mathrm{11...1}}}2,b=\underset{n}{\underbrace{\mathrm{11...1}}}4.$$

Vì $$\underset{n}{\underbrace{\mathrm{11...1}}}4-\underset{n}{\underbrace{\mathrm{11...1}}}2=2$$ nên $b-a=2$ hay $b=a+2$.

Do đó $ab+1=a\left(a+2\right)+1=a^2+2a+1$

$={\left(a+1\right)}^2={\left(\underset{n}{\underbrace{\mathrm{11....1}}}2+1\right)}^2={\left(\underset{n}{\underbrace{\mathrm{11....1}}}3\right)}^2.$

Vậy $$ab+1$$ là số chính phương.