Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

Ta có ${{x}^{4}}-4{{x}^{3}}+8{{x}^{2}}-16x+16=0$

$\Leftrightarrow ({{x}^{4}}+8{{x}^{2}}+16)-(4{{x}^{3}}+16x)=0$

$\Leftrightarrow {{({{x}^{2}}+4)}^{2}}-4x({{x}^{2}}+4)=0$

$\Leftrightarrow ({{x}^{2}}+4)({{x}^{2}}+4-4x)=0\Leftrightarrow ({{x}^{2}}+4){{(x-2)}^{2}}=0$

$\Leftrightarrow \left[ \begin{matrix} {{x}^{2}}+4=0 \\ {{(x-2)}^{2}}=0 \\ \end{matrix} \right.\Leftrightarrow \left[ \begin{matrix} {{x}^{2}}=-4(L) \\ x-2=0 \\ \end{matrix} \right.\Leftrightarrow x=2$ .

Vậy ${{x}_{0}}=2$ .

Ta có

$\begin{array}{l} {{x}^{4}}+16 \\ ={{x}^{4}}+8{{x}^{2}}+16-8{{x}^{2}} \\ ={{\left( {{x}^{2}}+4 \right)}^{2}}-{{\left( x\sqrt{8} \right)}^{2}} \\ =\left( {{x}^{2}}-x\sqrt{8}+4 \right)\left( {{x}^{2}}+x\sqrt{8}+4 \right) \end{array}$

Ta có

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

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

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

Ta có $3{{x}^{2}}+8x+5=0$

$\Leftrightarrow 3{{x}^{2}}+3x+5x+5=0\Leftrightarrow 3x(x+1)+5(x+1)=0\Leftrightarrow (3x+5)(x+1)=0$

$\Leftrightarrow \left[ \begin{matrix} 3x+5=0 \\ x+1=0 \\ \end{matrix} \right.\Leftrightarrow \left[ \begin{matrix} x=-\dfrac{5}{3} \\ x=-1 \\ \end{matrix} \right.$ .

Vậy $x=-\dfrac{5}{3};x=-1$ .

Ta có

$(I):4{{x}^{2}}+4x-9{{y}^{2}}+1=(4{{x}^{2}}+4x+1)-9{{y}^{2}}={{(2x+1)}^{2}}-{{(3y)}^{2}}=(2x+1+3y)(2x+1-3y)$ nên $(I)$ đúng.

Và $(II):5{{x}^{2}}-10xy+5{{y}^{2}}-20{{z}^{2}}=5({{x}^{2}}-2xy+{{y}^{2}}-4{{z}^{2}})=5[{{(x-y)}^{2}}-{{(2z)}^{2}}]$

= 5(x - y - 2z)(x - y + 2z) nên $(II)$ sai.

Ta có $6{{x}^{3}}+{{x}^{2}}-2x=0\Leftrightarrow x(6{{x}^{2}}+x-2)=0\Leftrightarrow x(6{{x}^{2}}+4x-3x-2)=0$

$\Leftrightarrow x[2x(3x+2)-(3x+2)]=0\Leftrightarrow x(3x+2)(2x-1)=0$

$\Rightarrow x=0$ hoặc $3x+2=0$ hoặc $2x-1=0$

Suy ra $x=0;x=-\dfrac{2}{3};x=\dfrac{1}{2}$ .

Vậy giá trị nhỏ nhất cần tìm là $x=-\dfrac{2}{3}$ .

Ta có ${{x}^{4}}+4{{x}^{2}}-5={{x}^{4}}-{{x}^{2}}+5{{x}^{2}}-5={{x}^{2}}({{x}^{2}}-1)+5({{x}^{2}}-1)=({{x}^{2}}+5)({{x}^{2}}-1)$

$=({{x}^{2}}+5)(x-1)(x+1)$ .

+ ${{x}^{2}}+5x+4={{x}^{2}}+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)$ .

+ ${{x}^{2}}-9x+8={{x}^{2}}-x-8x+8=x(x-1)-8(x-1)=(x-1)(x-8)$ .

+ ${{x}^{2}}+x-6={{x}^{2}}-2x+3x-6=x(x-2)+3(x-2)=(x-2)(x+3)$ .

Ta có

$\begin{array}{l} Pt\Leftrightarrow \left( 4{{x}^{2}}-4x+1 \right)-4=0 \\ \Leftrightarrow {{\left( 2x-1 \right)}^{2}}-{{2}^{2}}=0 \\ \Leftrightarrow \left( 2x-3 \right)\left( 2x+1 \right)=0 \\ \Leftrightarrow \left[ \begin{array}{l} x=\dfrac{3}{2} \\ x=-\dfrac{1}{2} \end{array} \right. \end{array}$

Vậy phương trình đã cho không có nghiệm nguyên.

Ta có

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

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

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

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

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

$=\left( x+y \right)\left( x+y \right)\left( x+y-3y \right)$

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

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

$=(x+y)[{{x}^{2}}-2xy+{{y}^{2}}]=(x+y){{(x-y)}^{2}}$ .

Vì $x=y\Leftrightarrow x-y=0$ nên $D=(x+y){{(x-y)}^{2}}=0$ .

Ta có

$\begin{array}{l} {{x}^{2}}+5x-6={{x}^{2}}-x+6x-6 \\ =\left( {{x}^{2}}-x \right)+6\left( x-1 \right) \\ =x\left( x-1 \right)+6\left( x-1 \right) \\ =\left( x-1 \right)\left( x+6 \right) \end{array}$

Ta có $25-{{a}^{2}}+2ab-{{b}^{2}}=25-({{a}^{2}}-2ab+{{b}^{2}})={{5}^{2}}-{{(a-b)}^{2}}=(5+a-b)(5-a+b)$ .

Ta có $ab(a-b)+bc(b-c)+ca(c-a)=ab(a-b)+bc[b-a+a-c\text{ }\!\!]\!\!\text{ }+ac(c-a)$

$=ab(a-b)-bc(a-b)+bc(a-c)-ac(a-c)=(a-b)(ab-bc)+(a-c)(bc-ac)$

$=b(a-b)(a-c)-c(a-c)(a-b)=(a-b)(a-c)(b-c)$ .

Ta có

$5{{x}^{3}}z-10{{x}^{2}}z-5x{{z}^{3}}-5x{{y}^{2}}z+5xz+10xy{{z}^{2}}$

$=5xz\left( {{x}^{2}}-2x-{{z}^{2}}-{{y}^{2}}+1+2yz \right)$

$=5xz\left[ \left( {{x}^{2}}-2x+1 \right)-\left( {{y}^{2}}-2yz+{{z}^{2}} \right) \right]$

$=5xz\left[ {{\left( x-1 \right)}^{2}}-{{\left( y-z \right)}^{2}} \right]$

$=5xz\left( x-1-y+z \right)\left( x-1+y-z \right)$

Ta có ${{x}^{2}}-7xy+10{{y}^{2}}={{x}^{2}}-2xy-5xy+10{{y}^{2}}=({{x}^{2}}-2xy)-(5xy-10{{y}^{2}})$

$=x(x-2y)-5y(x-2y)=(x-2y)(x-5y)$ .

Vậy ta cần điền $x-5y$ .

Ta có $C=xyz-(xy+yz+zx)+x+y+z-1$

$=(xyz-xy)-(yz-y)-(zx-x)+(z-1)$

$=xy(z-1)-y(z-1)-x(z-1)+(z-1)=(z-1)(xy-y-x+1)$

$=(z-1).\text{ }\!\![\!\!\text{ }y(x-1)-(x-1)\text{ }\!\!]\!\!\text{ }=(z-1)(y-1)(x-1)$

Với $x=9;y=10;z=101$ , ta có:

$C=(101-1)(10-1)(9-1)=100.9.8=7200$ .

Ta có $A={{x}^{2}}-4{{y}^{2}}+4x+4=({{x}^{2}}+4x+4)-4{{y}^{2}}$

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

Thay $x=62,y=-18$ ta được:

$A=(62+2-2.(-18))(62+2+2.(-18))=100.28=2800$ .

Ta có

$\begin{array}{l} 5{{x}^{2}}-10xy+5{{y}^{2}}-20{{z}^{2}} \\ =5\left( {{x}^{2}}-2xy+{{y}^{2}}-4{{z}^{2}} \right) \end{array}$

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

$=5\left( x-y+2z \right)\left( x-y-2z \right)$

Ta có ${{x}^{4}}+64={{({{x}^{2}})}^{2}}+16{{x}^{2}}+64-16{{x}^{2}}={{({{x}^{2}})}^{2}}+2.8.x+{{8}^{2}}-{{(4x)}^{2}}$

$={{({{x}^{2}}+8)}^{2}}-{{(4x)}^{2}}$ .

Ta có $3{{x}^{2}}-5x-2=3{{x}^{2}}+x-6x-2=x(3x+1)-2(3x+1)=(x-2)(3x+1)$ .

+ ${{x}^{2}}+5x+4={{x}^{2}}+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)$ .

+ ${{x}^{2}}-9x+8={{x}^{2}}-x-8x+8=x(x-1)-8(x-1)=(x-1)(x-8)$ .

+ ${{x}^{2}}+x-6={{x}^{2}}+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2)$ .

Ta có $3{{x}^{2}}+13x+10=0$

$\Leftrightarrow 3{{x}^{2}}+3x+10x+10=0\Leftrightarrow 3x(x+1)+10(x+1)=0$

$\Leftrightarrow (x+1)(3x+10)=0\Leftrightarrow \left[ \begin{matrix} x+1=0 \\ 3x+10=0 \\ \end{matrix} \right.\Leftrightarrow \left[ \begin{matrix} x=-1 \\ x=-\dfrac{10}{3} \\ \end{matrix} \right.$ .

$\Rightarrow 2{{x}_{1}}{{x}_{2}}=2.(-1).\left( -\dfrac{10}{3} \right)=\dfrac{20}{3}$ .

Ta có ${{x}^{2}}-6x+8={{x}^{2}}-4x-2x+8=x(x-4)-2(x-4)=(x-4)(x-2)$ .

Ta có ${{({{x}^{2}}+x)}^{2}}+4{{x}^{2}}+4x-12={{({{x}^{2}}+x)}^{2}}+4({{x}^{2}}+x)-12$ .

Đặt $t={{x}^{2}}+x$ ta được

${{t}^{2}}+4t-12={{t}^{2}}+6t-2t-12=t(t+6)-2(t+6)=(t-2)(t+6)=({{x}^{2}}+x-2)({{x}^{2}}+x+6)$

Vậy số cần tìm là $6$ .

$\begin{array}{l} 16x-5{{x}^{2}}-3 \\ =15x-5{{x}^{2}}-3+x \\ =\left( 15x-5{{x}^{2}} \right)-\left( 3-x \right) \end{array}$

$\begin{array}{l} =5x\left( 3-x \right)-\left( 3-x \right) \\ =\left( 3-x \right)\left( 5x-1 \right) \end{array}$