Phân tích đa thức thành nhân tử (P2)

Ta có

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

Khi đó với $x-y=2z$ thì $B=0$

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

$=(8{{x}^{2}}+8x).(-6)=8.x.(x+1).(-6)$

$=-48x(x+1)$ nên $m=-48 < 0$ .

Ta có $5{{x}^{2}}-10x+5=0$

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

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

$\Leftrightarrow {{(x-1)}^{2}}=0$

$\Leftrightarrow x-1=0$

$\Leftrightarrow x=1$ .

Vậy $x=1$ .

Ta có $4{{b}^{2}}{{c}^{2}}-{{({{c}^{2}}+{{b}^{2}}-{{a}^{2}})}^{2}}={{(2bc)}^{2}}-{{({{c}^{2}}+{{b}^{2}}-{{a}^{2}})}^{2}}=(2bc+{{c}^{2}}+{{b}^{2}}-{{a}^{2}})(2bc-{{c}^{2}}-{{b}^{2}}+{{a}^{2}})$

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

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

$\begin{array}{l} 8-27{{a}^{3}}{{b}^{6}} \\ ={{2}^{3}}-{{\left( 3a{{b}^{2}} \right)}^{3}} \\ =\left( 23a{{b}^{2}} \right)\left( 4+6a{{b}^{2}}+9{{a}^{2}}{{b}^{4}} \right) \end{array}$

Ta có ${{(2x-1)}^{2}}-{{(5x-5)}^{2}}=0$

$\Leftrightarrow (2x-1+5x-5)(2x-1-5x+5)=0$

$\Leftrightarrow (7x-6)(4-3x)=0\Leftrightarrow \left[ \begin{matrix} 7x-6=0 \\ 4-3x=0 \\ \end{matrix} \right.$

$\Leftrightarrow \left[ \begin{matrix} x=\dfrac{6}{7} \\ x=\dfrac{4}{3} \\ \end{matrix} \right.$ .

Vậy có hai giá trị của $x$ thỏa mãn yêu cầu.

Ta có $4{{(3x-5)}^{2}}-9{{(9{{x}^{2}}-25)}^{2}}=0\Leftrightarrow 4.{{(3x-5)}^{2}}-9{{[{{(3x)}^{2}}-{{5}^{2}}]}^{2}}=0$

$\Leftrightarrow 4{{(3x-5)}^{2}}-9{{[(3x-5)(3x+5)\text{ }\!\!]\!\!\text{ }}^{2}}=0\Leftrightarrow 4{{(3x-5)}^{2}}-9{{(3x-5)}^{2}}{{(3x+5)}^{2}}=0$

$\Leftrightarrow {{(3x-5)}^{2}}\text{ }\!\![\!\!\text{ }4-9{{(3x+5)}^{2}}\text{ }\!\!]\!\!\text{ }=0$

$\Leftrightarrow {{(3x-5)}^{2}}\text{ }\!\![\!\!\text{ }4-{{(3(3x+5))}^{2}}\text{ }\!\!]\!\!\text{ }=0\Leftrightarrow {{(3x-5)}^{2}}({{2}^{2}}-{{(9x+15)}^{2}})=0$

$\Leftrightarrow {{(3x-5)}^{2}}(2+9x+15)(2-9x-15)=0\Leftrightarrow {{(3x-5)}^{2}}(9x+17)(-9x-13)=0$

$\Leftrightarrow \left[ \begin{matrix} 3x-5=0 \\ 9x+17=0 \\ -9x-13=0 \\ \end{matrix} \right.\Leftrightarrow \left[ \begin{matrix} x=\dfrac{5}{3} \\ x-\dfrac{17}{9} \\ x=-\dfrac{13}{9} \\ \end{matrix} \right.$ suy ra ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=\dfrac{5}{3}+\dfrac{-17}{9}+\dfrac{-13}{9}=-\dfrac{5}{3}$ .

Ta có $\dfrac{{{x}^{3}}}{8}+8{{y}^{3}}={{\left( \dfrac{x}{2} \right)}^{3}}+{{(2y)}^{3}}=\left( \dfrac{x}{2}+2y \right)\left[ {{\left( \dfrac{x}{2} \right)}^{2}}-\dfrac{x}{2}.2y+{{(2y)}^{2}} \right]$

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

$\begin{array}{l} {{x}^{2}}-10xy=4-25{{y}^{2}} \\ \Leftrightarrow {{x}^{2}}-2.x.5y+25{{y}^{2}}=4 \\ \Leftrightarrow {{\left( x-5y \right)}^{2}}=4 \\ \Leftrightarrow \left[ \begin{array}{l} x-5y=2 \\ x-5y=-2\end{array} \right. \end{array}$

Suy ra, $\left[ \begin{array}{l} 10y-2x=-4 \\ 10y-2x=4 \end{array} \right.$

Ta có

$\begin{array}{l} {{x}^{6}}-64{{y}^{6}}={{\left( {{x}^{3}} \right)}^{2}}-{{\left( 8{{y}^{3}} \right)}^{2}} \\ =\left( {{x}^{3}}-8{{y}^{3}} \right)\left( {{x}^{3}}+8{{y}^{3}} \right) \\ =\left( x-2y \right)\left( {{x}^{2}}+2xy+4{{y}^{2}} \right)\left( x+2y \right)\left( {{x}^{2}}-2xy+4{{y}^{2}} \right) \end{array}$

Ta có ${{(2x-5)}^{2}}-4{{(x-2)}^{2}}=0\Leftrightarrow {{(2x-5)}^{2}}-{{\text{ }\!\![\!\!\text{ }2(x-2)\text{ }\!\!]\!\!\text{ }}^{2}}=0\Leftrightarrow {{(2x-5)}^{2}}-{{(2x-4)}^{2}}=0$

$\Leftrightarrow (2x-5+2x-4)(2x-5-2x+4)=0\Leftrightarrow (4x-9)(-1)=0$

$\Leftrightarrow -4x+9=0$

$\Leftrightarrow 4x=9$

$\Leftrightarrow x=\dfrac{9}{4}$ .

Vậy $x=\dfrac{9}{4}$ .

Ta có $8{{x}^{3}}-64={{(2x)}^{3}}-{{4}^{3}}=(2x-4)(4{{x}^{2}}+8x+16)$ .

Ta có ${{(4{{x}^{2}}+2x-18)}^{2}}-{{(4{{x}^{2}}+2x)}^{2}}=(4{{x}^{2}}+2x-18+4{{x}^{2}}+2x)(4{{x}^{2}}+2x-18-4{{x}^{2}}-2x)$

$=(8{{x}^{2}}+4x-18)(-18)=2(4{{x}^{2}}+2x-9)(-18)=(-36)(4{{x}^{2}}+2x-18)\Rightarrow m=-36$ .

Ta có ${{(5x-4)}^{2}}-49{{x}^{2}}={{(5x-4)}^{2}}-{{(7x)}^{2}}=(5x-4+7x)(5x-4-7x)=(12x-4)(-2x-4)$

$=4(3x-1).(-2)(x+2)=-8(3x-1)(x+2)$ .

Ta có

$9{{\left( x+5 \right)}^{2}}-{{\left( x+7 \right)}^{2}}=0$

$\Leftrightarrow {{\left[ 3\left( x+5 \right) \right]}^{2}}-{{\left( x+7 \right)}^{2}}=0$

$\Leftrightarrow \left[ 3\left( x+5 \right)+x+7 \right]\left[ 3\left( x+5 \right)-\left( x+7 \right) \right]=0$

$\Leftrightarrow \left( 4x+22 \right)\left( 2x+8 \right)=0$

$\begin{array}{l} \Leftrightarrow 4\left( 2x+11 \right)\left( x+4 \right)=0 \\ \Leftrightarrow x=-\dfrac{11}{2};x=-4 \end{array}$

Ta có

$\begin{array}{l} 25{{x}^{4}}-10{{x}^{2}}y+{{y}^{2}} \\ ={{\left( 5{{x}^{2}} \right)}^{2}}-2.5{{x}^{2}}.y+{{y}^{2}} \\ ={{\left( 5{{x}^{2}}-y \right)}^{2}} \end{array}$

Ta có

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

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

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

$=2y(3{{x}^{2}}+{{y}^{2}})\Rightarrow A=2;B=3;C=1$ .

Suy ra $A+B+C=2+3+1=6$ .

Ta có ${{x}^{3}}{{y}^{3}}+6{{x}^{2}}{{y}^{2}}+12xy+8={{(xy)}^{3}}+3{{(xy)}^{2}}.2+3.xy{{.2}^{2}}+{{2}^{3}}={{(xy+2)}^{3}}$

Ta có

$\begin{array}{*{20}{l}} {{{\left( {x + y} \right)}^2} - {{\left( {x - y} \right)}^2} = 0}\\ { \Leftrightarrow \left[ {\left( {x + y} \right) + \left( {x - y} \right)} \right]\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right] = 0}\\ \begin{array}{l}  \Leftrightarrow 2x.2y = 0\\  \Leftrightarrow \left[ {\begin{array}{*{20}{l}} {x = 0}\\ {y = 0} \end{array}} \right. \end{array} \end{array}$

Vậy có vô số cặp thỏa mãn có dạng $\left( 0;y \right),\left( x;0 \right)$

Ta có

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

Ta có $9{{a}^{2}}-{{(a-3b)}^{2}}={{(3a)}^{2}}-{{(a-3b)}^{2}}=(3a+a-3b)(3a-a+3b)=(4a-3b)(2a+3b)$

Suy ra $m=2;n=3$ .

Ta có:

+) $4{{x}^{2}}+4x+1={{(2x)}^{2}}+2.2x.1+{{1}^{2}}={{(2x+1)}^{2}}$

+) $9{{x}^{2}}-24xy+16{{y}^{2}}={{(3x)}^{2}}-2.3x.4y+{{(4y)}^{2}}={{(3x-4y)}^{2}}$

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

Ta có

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

Vậy có 3 giá trị x thỏa mãn.