Chủ đề 1: Nhân đa, đơn thức có đáp án

Ta có: $\left(x-1\right)\left(x^2-x+1\right)-\left(x-x^2\right)\left(2x\right)$

                $=x^2\left(x-1\right)-x\left(x-1\right)+1\left(x-1\right)-\left(x.2x-x^2.2x\right)$

                $=\left(x.x^2-1.x^2\right)-\left(x.x-1.x\right)+x-1-\left(2x^2-2x^3\right)$

                $=x^3-x^2-x^2+x+x-1-2x^2+2x^3$

                $=3x^3-4x^2+2x-1$

Ta có: $\left(x-1\right)\left(2x-1\right)\left(3x-1\right)$

                $=\left(2x\left(x-1\right)-1\left(x-1\right)\right)\left(3x-1\right)$

                $=\left(2x^2-2x-x+1\right)\left(3x-1\right)$

                $=\left(2x^2-3x+1\right)\left(3x-1\right)$

                $=2x^2\left(3x-1\right)-3x\left(3x-1\right)+1\left(3x-1\right)$

                $=6x^3-2x^2-9x^2+3x+3x-1=6x^3-11x^2+6x-1$

Ta có: $\left(x\right)\left(x-y\right)-\left(x+1\right)\left(x+yx\right)+\left(y\right)\left(x-x{y}^2\right)$

                  $=\left(x.x-y.x\right)-\left(x.\left(x+1\right)+yx.\left(x+1\right)\right)+\left(x.y-x{y}^2.y\right)$

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

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

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

Ta có: $\left(\frac{2x^2+y^2}{2}\right)\left(\frac{x}{3}+\frac{y}{2}\right)\left(\frac{1-xy}{3}\right)$

                   $=\left(\frac{2x^2}{2}+\frac{y^2}{2}\right)\left(\frac{x}{3}+\frac{y}{2}\right)\left(\frac{1}{3}-\frac{xy}{3}\right)$

                   $=\left(\frac{2x^2}{2}\left(\frac{x}{3}+\frac{y}{2}\right)+\frac{y^2}{2}\left(\frac{x}{3}+\frac{y}{2}\right)\right)\left(\frac{1}{3}-\frac{xy}{3}\right)$

                   $=\left(\left(\frac{x}{3}.\frac{2x^2}{2}+\frac{y}{2}.\frac{2x^2}{2}\right)+\left(\frac{x}{3}.\frac{y^2}{2}+\frac{y}{2}.\frac{y^2}{2}\right)\right)\left(\frac{1}{3}-\frac{xy}{3}\right)$

                   $=\left(\frac{2x^3}{6}+\frac{x^{2}y}{2}+\frac{x{y}^2}{6}+\frac{y^3}{4}\right)\left(\frac{1}{3}-\frac{xy}{3}\right)$

                   $=\left(\frac{x^3}{3}\left(\frac{1}{3}-\frac{xy}{3}\right)+\frac{x^{2}y}{2}\left(\frac{1}{3}-\frac{xy}{3}\right)+\frac{x{y}^2}{6}\left(\frac{1}{3}-\frac{xy}{3}\right)+\frac{y^3}{4}\left(\frac{1}{3}-\frac{xy}{3}\right)\right)$

              $=\left(\frac{1}{3}\frac{x^3}{3}-\frac{xy}{3}\frac{x^3}{3}\right)+\left(\frac{1}{3}\frac{x^{2}y}{2}-\frac{xy}{3}\frac{x^{2}y}{2}\right)+\left(\frac{1}{3}\frac{x{y}^2}{6}-\frac{xy}{3}\frac{x{y}^2}{6}\right)+\left(\frac{1}{3}\frac{y^3}{4}-\frac{xy}{3}\frac{y^3}{4}\right)$

                   $=\left(\frac{x^3}{9}-\frac{2x^{4}y}{9}\right)+\left(\frac{x^{2}y}{6}-\frac{x^3{y}^3}{6}\right)+\left(\frac{x{y}^2}{18}-\frac{x^2{y}^3}{18}\right)+\left(\frac{y^3}{12}-\frac{x{y}^4}{12}\right)$

                   $=\frac{x^3-2x^{4}y}{9}+\frac{x^{2}y-x^3{y}^3}{6}+\frac{x{y}^2-x^2{y}^3}{18}+\frac{y^3-x{y}^4}{12}$

Ta có: $A=2x^2\left(x^2-5\right)-x^2\left(x-3x^2\right)-\left(x^2-x^4\right)$

                   $=2x^4-10x^2-x^3+3x^4-x^2+x^4$

                   $=-6x^2-x^3-11x^2=-x^3-11x^2$

         Tại x = 1 thay vào ta được: $A={6.1}^4-1^3-{11.1}^2=6-1-11=-6$

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

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

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

Áp dụng hằng đẳng thức ta có:

                   ${\left(x^2+y\right)}^2={\left(x^2\right)}^2+2\left(x^2\right)\left(y\right)+{\left(y\right)}^2=x^4+2x^{2}y+y^2$

Áp dụng hằng đẳng thức ta có:

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

Áp dụng hằng đẳng thức ta có:

        $16x^4-y^4={\left(4x^2\right)}^2-{\left(y^2\right)}^2=\left(4x^2-y^2\right)\left(4x^2+y^2\right)$

Áp dụng hằng đẳng thứ ta có:

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

                         $=x^2+2xy+y^2+2xz+2yz+z^2$

                         $=x^2+y^2+z^2+2xy+2xz+2yz$

Áp dụng bất đẳng thức ta được:

                   $-x^3-8y^3={\left(-x\right)}^3-{\left(2y\right)}^3$

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

               $=\left(-x-2y\right)\left(x^2-xy+4y^2\right)$

Áp dụng bất đẳng thức ta có:

               ${\left(y-x\right)}^3+{\left(x+y\right)}^3+{\left(x-y\right)}^3-\left(x^3+y^3\right)$

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

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

Ta có: $x^{n+1}\left(x+y\right)-y\left(x^{n+1}+y^{n+1}\right)$

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

Ta có: $\left(-3x^n{y}^n+\frac{1}{2}{x}^n{y}^{2n}-3\right)\left(-2x^{2n}{y}^n\right)+x^{2n}{y}^n\left(x^n{y}^{2n}+1\right)$

           $=\left(-3x^n{y}^n\right).\left(-2x^{2n}{y}^n\right)+\left(\frac{1}{2}{x}^n{y}^{2n}\right).\left(-2x^{2n}{y}^n\right)-3\left(-2x^{2n}{y}^n\right)+x^{2n}{y}^n\left(x^n{y}^{2n}+1\right)$

                   $=6x^{3n}{y}^{2n}-x^{3n}{y}^{3n}+6x^{2n}{y}^n+x^{3n}{y}^{3n}+x^{3n}{y}^{3n}+x^{2n}{y}^n$

                   $=6x^{3n}{y}^{2n}+7x^{2n}{y}^n=x^{2n}{y}^n\left(6x^n{y}^n+7\right)$

Ta có: $x^2\left(a-b\right)+b\left(1-x\right)+x\left(bx+b\right)-ax\left(x+1\right)$

Ta có: $3a^2\left(4a-3b\right)-3a^2\left(3a+b\right)-3a^3$

                   $=12a^3-9a^{2}b-9a^3-3a^{2}b-3a^3=-12a^{2}b$

Ta có: $4\left(x-2y\right)\left(x+y\right)+4{\left(x+y\right)}^2+{\left(x-2y\right)}^2-12xy+4y^4$

                   $=\left(4x-8y\right)\left(x+y\right)+4\left(x^2+2xy+y^2\right)+x^2-4xy+4y^2-12xy+4y^2$

                   $=4x\left(x+y\right)-8y\left(x+y\right)+4x^2+8xy+4y^2+x^2-16xy+8y^2$

                   $=4x^2+4xy-8xy-8y^2+5x^2-8xy+12y^2$

                   $=9x^2-12xy+4y^2={\left(3x\right)}^2-2.3x.2y+{\left(2y\right)}^2={\left(3x-2y\right)}^2$

Ta có: $\left(-a^{4n}{x}^{5n}\right)\left(-a^{6n}{x}^n+2a^{3n}{x}^{2n}-11a{x}^{5n}\right)-a^{5n}{x}^{5n}\left(a^{5n}{x}^n+11x^{5n}\right)$

Ta có: $\left(x^n-y^n\right)\left(x^n+y^n\right)\left(x^{2n}-2x^n{y}^n+2y^{2n}\right)$

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

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

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

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

Ta có: $\left(x^n+2\right)\left(2-x^n\right)\left(4+x^{2n}\right)$

                   $=\left(2x^n+4-x^{2n}-2x^n\right)\left(4+x^{2n}\right)=\left(4-x^{2n}\right)\left(4+x^{2n}\right)$

Ta có:

         VT $=\left(x^2-3\right)\left(x+4\right)+\left(x-x^2\right)\left(x+3\right)$

              $=x^3-3x+4x^2-12+x^2-x^3+3x-3x^2=2x^2-12=$ VP

         $\Rightarrow$ đpcm.

Ta có:

         VT $=\left(x^2-3\right)\left(x+4\right)+\left(x-x^2\right)\left(x+3\right)$

              $=x^3-3x+4x^2-12+x^2-x^3+3x-3x^2=2x^2-12=$ VP

         $\Rightarrow$ đpcm.

Ta có:

         VT $=\left(4-x+y\right)\left(4+x-y\right)+\left(x-y\right)\left(x+y\right)$

              $=16-4x+4y+4x-x^2+yx-4y+xy-y^2+x^2-yx+xy-y^2$

              $=16+2xy-2y^2=16+2y\left(x-y\right)=$ VP

         $\Rightarrow$ đpcm.

Ta có:

         VT $=\left(x^2+2-2x\right)\left(x^2+2+2x\right)$

              $=x^4+2x^2-4x^3+2x^2+4-4x+4x^3+4x-4x^2$

              $=x^4+4=$ VP

         $\Rightarrow$ đpcm.

Ta có: $A=\left(x-4\right)\left(2x+5\right)-2x\left(x-3\right)-\left(3x-8\right)$

                   $=2x^2-8x+5x-20-2x^2+6x-3x+8=-12$

Vậy giá trị của biếu thức A không phụ thuộc vào biến x

Ta có: $B=\left(x^2+2x+3\right)\left(3x^2-2x+1\right)-3x^2\left(x^2+2\right)-4x\left(x^2-1\right)$

                   $=3x^4+6x^3+9x^2-2x^3-4x^2-6x+x^2+2x+3-3x^4-6x^2-4x^3+4x$

                   $=3$

Vậy giá trị của biểu thức B không phụ thuộc vào biến x

Ta có: $C=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)$

                   $=6x^2-10x+33x-55-6x^2-9x-14x-21=-76$

Vậy giá trị của biểu thức C không phụ thuộc vào biến x

Ta có: $y^2-6y+10=y^2-2.y.3+9+1={\left(y-3\right)}^3+1$

        Ta thấy ${\left(y-3\right)}^2\ge 0$ suy ra ${\left(y-3\right)}^2+1>0$ $\forall x\Rightarrow$ đpcm

Ta có: $9x^2-6x+25y^2+10y+4$

                   $={\left(3x\right)}^2-2.3x+1+{\left(5y\right)}^2+2.5y+1+3$

                   $={\left(3x-1\right)}^2+{\left(5y+1\right)}^2+3>0$  $\forall x\Rightarrow$ đpcm

Ta có: $x^2-x+\frac{3}{2}=x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}+\frac{3}{2}$

                   $={\left(x-\frac{1}{2}\right)}^2+\frac{5}{4}>0$ $\forall x\Rightarrow$ đpcm

Ta có: $x^2+y^2+z^2-4x-4y-4z+12$

                   $=x^2-4x+4+y^2-4y+4+z^2-4z+4$

                   $={\left(x-2\right)}^2+{\left(y-2\right)}^2+{\left(z-2\right)}^2\ge 0$ với mọi số thực x, y, z

Dấu “=” xảy ra khi và chỉ khi $x=y=z=2$.

Ta có: $A=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1$

              $A=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1$

Đặt $t=x^2+5x+4$, khi đó ta có:

              $A=t\left(t+2\right)+1=t^2+2t+1={\left(t+1\right)}^2$

Vậy $A={\left(x^2+5x+5\right)}^2$ (đpcm)

Ta có: $A=4x-2x^2+3=-2x^2+4x-2+5$

                   $=-2\left(x^2-2x+1\right)+5=5-2{\left(x-1\right)}^2\le 5$

Vậy giá trị lớn nhất của biểu thức A là 5 khi $x-1=0\iff x=1$.

Ta có: $B=\left(1-x\right)\left(2+x\right)+\left(3-x\right)\left(4+x\right)$

                   $=2-2x+x-x^2+12-4x+3x-x^2$

                   $=-2x^2-2x+14=-2\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{1}{2}+14$

                   $=\frac{27}{2}-2{\left(x+\frac{1}{2}\right)}^2\le \frac{27}{2}$

Vậy giá trị lớn nhất của biểu thức B là $\frac{27}{2}$ khi $x+\frac{1}{2}=0\iff x=-\frac{1}{2}$.

Ta có: $C=5-8x-x^2=-x^2-8x-16+16+5$

                   $=-\left(x^2+8x+16\right)+21=-{\left(x+4\right)}^2+21$

                   $=21-{\left(x+4\right)}^2\le 21$ $\forall x$

Vậy giá trị lớn nhất của C là 21 khi $x+4=0\Rightarrow x=-4$

Ta có: $D=-3x\left(x+3\right)-7=-3x^2-9x-7$

                   $=-3\left(x^2+2x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)-7=-3{\left(x-\frac{3}{2}\right)}^2+\frac{27}{4}-7$

                   $=-3{\left(x+\frac{3}{2}\right)}^2-\frac{1}{4}$

                   $=\left(-\frac{1}{4}\right)-3{\left(x+\frac{3}{2}\right)}^2\le -\frac{1}{4}$ $\forall x$

Vậy giá trị lớn nhất của D là $-\frac{1}{4}$ khi $x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}$

Ta có: $A=4x^2-4x+2=4x^2-4x+1+1={\left(2x-1\right)}^2+1\ge 1$

Vậy giá trị nhỏ nhất của biểu thức A là 1 khi $2x-1=0\iff x=\frac{1}{2}$.

Ta có: $B=12+6x^2+9x=3\left(x^2+3x+4\right)$

                   $=3\left(x^2+2\frac{3}{2}x+\frac{9}{4}+\frac{7}{4}\right)=3{\left(x+\frac{3}{2}\right)}^2+\frac{21}{4}\ge \frac{21}{4}$

Vậy giá trị nhỏ nhất của biểu thức B là $\frac{21}{4}$ khi $x+\frac{3}{2}=0\iff x=-\frac{3}{2}$.

Ta có: $A=x\left(x+3\right)\left(x-1\right)\left(x-4\right)=x\left(x-1\right)\left(x+3\right)\left(x-4\right)$

                   $=\left(x^2-x\right)\left(x^2+3x-4x-12\right)=\left(x^2-x\right)\left(x^2-x-12\right)$

Đặt $t=\left(x^2-x\right)$ thay vào ta được:

                   $A=t\left(t-12\right)=t^2-12t+36-36={\left(t-6\right)}^2-36\ge -36$

Giá trị nhỏ nhất của $A=-36$ khi và chỉ khi $t-6=0\iff t=6$

                   $\iff x^2-x=6\iff x^2+2x-3x-6=0$

                   $\iff x\left(x+2\right)-3\left(x+2\right)=0$

                   $\iff \left(x-3\right)\left(x+2\right)=0\iff \left[\begin{array}{l}x=3\\ x=-2\end{array}\right.$

Vậy giá trị nhỏ nhất của $A=-36$ khi và chỉ khi $x=3$ hoặc $x=-2$

Ta có: $B=4x^4+4x^3+5x^2+4x+3$

                   $=4x^4+4x^3+x^2+4x^2+4x+1+2$

                   $=x^2\left(4x^2+4x+1\right)+\left(4x^2+4x+1\right)+2$

                   $=x^2{\left(2x+1\right)}^2+{\left(2x+1\right)}^2+2\ge 2$

Mặt khác $B=2\iff \left\{\begin{array}{l}\left[\begin{array}{l}{x}^2=0\\ 2x+1=0\end{array}\right.\\ 2x+1=0\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{l}x=0\\ x=-\frac{1}{2}\end{array}\right.\\ x=-\frac{1}{2}\end{array}\right.\iff x=-\frac{1}{2}$

Vậy giá trị nhỏ nhất của $B=2$ khi và chỉ khi $x=-\frac{1}{2}$

Ta có: $A=x^2+5x+8=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+8$

                   $={\left(x+\frac{5}{2}\right)}^2+\frac{7}{4}\ge \frac{7}{4}$ $\forall x$