Nhân đa thức với đa thức

$\begin{array}{*{35}{l}} \left( x+3 \right)\left( {{x}^{2}}+3x-5 \right) \\ =x.\left( {{x}^{2}}+3x-5 \right)+3.\left( {{x}^{2}}+3x-5 \right) \\ ={{x}^{3}}+3{{x}^{2}}-5x+3{{x}^{2}}+9x-15 \\ ={{x}^{3}}+6{{x}^{2}}+4x-15. \end{array}$

Khi đó $a+b+c=4+6-15=-5$

$\begin{array}{l} \left( xy-1 \right)\left( xy+x+5 \right) \\ =xy.\left( xy+x+5 \right)-\left( xy+x+5 \right) \\ ={{x}^{2}}{{y}^{2}}+{{x}^{2}}y+5xy-xy-x-5 \\ ={{x}^{2}}{{y}^{2}}+{{x}^{2}}y-x+4xy-5 \end{array}$

$\begin{array}{*{35}{l}} \left( {{x}^{2}}-2x+1 \right)\left( x-1 \right) \\ =x.\left( {{x}^{2}}-2x+1 \right)-\left( {{x}^{2}}-2x+1 \right) \\ ={{x}^{3}}-2{{x}^{2}}+x-{{x}^{2}}+2x-1 \\ ={{x}^{3}}-3{{x}^{2}}+3x-1 \end{array}$

Ta có $5(3x+5)-4(2x-3)=5x+3(2x-12)+1\Leftrightarrow 15x+25-8x+12=5x+6x-36+1$

$\Leftrightarrow 7x+37=11x-35\Leftrightarrow 4x=72\Leftrightarrow x=18$ .

Vậy $x=18$ .

Suy ra $17 < x < 19$ .

Ta có: $C=x(y+z)-y(z+x)-z(x-y)$

$=xy+xz-yz-xy-zx+zy=(xy-xy)+(zy-zy)+(xz-zx)=0$

Nên $C$ không phụ thuộc vào $x;y;z$ .

$\begin{array}{l} \left[ \left( {{x}^{2n}}+{{x}^{n}}\cdot {{y}^{n}}+{{y}^{2n}} \right)\left( {{x}^{n}}-{{y}^{n}} \right) \right]\cdot \left( {{x}^{3n}}+{{y}^{3n}} \right) \\ =\left( {{x}^{3n}}+{{x}^{2n}}{{y}^{n}}+{{x}^{n}}{{y}^{2n}}-{{x}^{2n}}{{y}^{n}}-{{x}^{n}}{{y}^{2n}}-{{y}^{3n}} \right)\left( {{x}^{3n}}+{{y}^{3n}} \right) \\ =\left( {{x}^{3n}}-{{y}^{3n}} \right)\left( {{x}^{3n}}+{{y}^{3n}} \right) \\ ={{x}^{6n}}+{{x}^{3n}}{{y}^{3n}}-{{x}^{3n}}{{y}^{3n}}-{{y}^{6n}} \\ ={{x}^{6n}}-{{y}^{6n}} \end{array}$

$\begin{array}{l} \left( xy-1\text{ } \right)\left( \text{ }{{x}^{3}}-2x-6 \right) =xy\left( \text{ }{{x}^{3}}-2x-6 \right)\text{ }+~~\left( -1 \right)\left( \text{ }{{x}^{3}}-2x-6 \right) \\ =xy.\text{ }{{x}^{3}}+xy.\left( -2x \right)+xy.\left( -6 \right)\text{ }+\left( -1 \right).\text{ }{{x}^{3}}+\left( -1 \right).\left( -\text{ }2x \right)+\left( -1 \right).\left( -6 \right) \\ ={{x}^{4}}y-{{x}^{3}}-2{{x}^{2}}y\text{ }+2x-6xy+6 \end{array}$

Ta có $A=x(x+1)+(1-x)(1+x)-x={{x}^{2}}+x+1+x-x-{{x}^{2}}-x=1$

Suy ra $A=1 > 0$ .

Ta có $P={{x}^{10}}-13{{x}^{9}}+13{{x}^{8}}-13{{x}^{7}}+...-13x+10$

$={{x}^{10}}-12{{x}^{9}}-{{x}^{9}}+12{{x}^{8}}+{{x}^{8}}-12{{x}^{7}}-{{x}^{7}}+12{{x}^{6}}+...+{{x}^{2}}-12x-x+10$

$={{x}^{9}}(x-12)-{{x}^{8}}(x-12)+{{x}^{7}}(x-12)-...+x(x-12)-x+10$

Thay $x=12$ vào $P$ ta được:

$P={{12}^{9}}.(12-12)-{{12}^{8}}(12-12)+{{12}^{7}}(12-12)-...+12(12-12)-12+10=0+...+0-2=-2$

Vậy $P=-2$ .

Ta có

$\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$

$\begin{array}{l} ={{a}^{3}}+a{{b}^{2}}+a{{c}^{2}}+{{a}^{2}}b-abc+{{a}^{2}}c-{{a}^{2}}b+{{b}^{3}}+b{{c}^{2}}- \\ a{{b}^{2}}+{{b}^{2}}c-abc-{{a}^{2}}c-{{b}^{2}}c+{{c}^{3}}-abc+b{{c}^{2}}-a{{c}^{2}} \\ ={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc \end{array}$

Ta có $M=x({{x}^{3}}+{{x}^{2}}-3x-2)-({{x}^{2}}-2)({{x}^{2}}+x-1)$

$=x.{{x}^{3}}+x.{{x}^{2}}-3x.x-2.x-({{x}^{2}}.{{x}^{2}}+{{x}^{2}}.x-{{x}^{2}}-2{{x}^{2}}-2x+2)$

$={{x}^{4}}+{{x}^{3}}-3{{x}^{2}}-2x-({{x}^{4}}+{{x}^{3}}-3{{x}^{2}}-2x+2)$

$={{x}^{4}}+{{x}^{3}}-3{{x}^{2}}-2x-{{x}^{4}}-{{x}^{3}}+3{{x}^{2}}+2x-2=-2$ .

Vậy $M=-2$ .

Ta có

$\begin{array}{l} \left( {{x}^{2}}{{y}^{2}}-xy+2y \right)\left( x-2y \right) \\ =x\left( {{x}^{2}}{{y}^{2}}-xy+2y \right)-2y\left( {{x}^{2}}{{y}^{2}}-xy+2y \right) \\ ={{x}^{3}}{{y}^{2}}-{{x}^{2}}y+2xy-2{{x}^{2}}{{y}^{3}}+2x{{y}^{2}}-4{{y}^{2}} \end{array}$

Ta có $D=x({{x}^{2n-1}}+y)-y(x+{{y}^{2n-1}})+{{y}^{2n}}-{{x}^{2n}}+5$

$=x.{{x}^{2n-1}}+x.y-y.x-y.{{y}^{2n-1}}+{{y}^{2n}}-{{x}^{2n}}+5$

$={{x}^{2n}}+xy-xy-{{y}^{2n}}+{{y}^{2n}}-{{x}^{2n}}+5=({{x}^{2n}}-{{x}^{2n}})+(xy-xy)+({{y}^{2n}}-{{y}^{2n}})+5$

$=0+0+0+5=5$ .

$A=(3x+7)(2x+3)-(3x-5)(2x+11)$

$=3x.2x+3x.3+7.2x+7.3-(3x.2x+3x.11-5.2x-5.11)$

$=2{{x}^{2}}+x-{{x}^{3}}-2{{x}^{2}}+{{x}^{3}}-x+3=3$

Gọi $x\,\,(x > 2)$ là độ dài đáy nhỏ của hình thang.

Theo giả thiết ta có độ dài đáy lớn là $2x$ , chiều cao của hình thang là $x-2$ .

Diện tích hình thang là $S=\dfrac{(x+2x)(x-2)}{2}=\dfrac{3x(x-2)}{2}=\dfrac{3{{x}^{2}}-6x}{2}$ (đvdt).