Phiếu bài tập tuần Toán 8 - Tuần 07

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

b) $\left( 2a-3b \right)\left( 4a-b \right)-\left( {{a}^{2}}-{{b}^{2}} \right)-{{\left( 3b-2a \right)}^{2}}$

c) ${{a}^{8}}-1$

d) ${{(x-y)}^{2}}+4(x-y)-12$

e) ${{x}^{2}}+{{y}^{2}}+3x-3y-2xy-10$

f) ${{x}^{2}}-6x-16\,$

g) $(x+2)(x+3)(x+4)(x+5)-24$

h) $({{x}^{2}}+6x+5)({{x}^{2}}+10x+21)+15$

$A\, = \, - 2{{\rm{x}}^2}\, + \;6{\rm{x}}\, + \,9$$\begin{array}{l}
 = \, - 2({x^2}\, - \,3{\rm{x)}}\,{\rm{ + }}\,{\rm{9}}\,\,{\rm{ = }}\,\,\,{\rm{ - 2}}\left( {{x^2}\, - \,2.\,x.\,\frac{3}{2}\, + \,\frac{9}{4}} \right)\,\, + \,\frac{9}{2}\, + \,9\\
 = \, - 2{\left( {x\, - \,\frac{3}{2}} \right)^2}\, + \,\frac{{27}}{2}\, \le \,\,\frac{{27}}{{2\,}}\,,\,\forall \,x
\end{array}$

Vì $ - 2{\left( {x\, - \,\frac{3}{2}} \right)^2}\, \le \,0\,$ nên  $A\,\, \le \,\,\frac{{27}}{2}$

Vậy A_max = $\frac{{27}}{2} \Leftrightarrow \,x\, = \,\frac{3}{2}$

$\begin{array}{l}
B\, = \,( - {{\rm{x}}^2}\, + \;2{\rm{x}}y\, - \,{y^2})\, + 4(x\, - \,y)\,\, + \,12x\, - \,4{x^2}\, - \,14\\
B\,\, = \,\, - [({{\rm{x}}^2}\, - \;2{\rm{x}}y\, + \,{y^2})\, - 4(x\, - \,y)\, + \,4]\, - (4{x^2}\, - \,12x + 9)\,\, - \,\,1
\end{array}$

$B\, = \, - [{(x\, - \,y)^2}\, - \,2.(x - \,y).2\, + \,{2^2}{\rm{]}}\, - \,{(2{\rm{x}}\, - 3)^2}\, - \,1$

$B\,\, = \,\, - {(x\, - \,y\, - \,2)^2}\, - \,{(2{\rm{x}}\, - \,3)^2}\, - \,1$

Vì $ - {(x\, - \,y\, - \,2)^2}\, \le \,0,\,\, - {(2{\rm{x}}\, - \,3)^2}\, \le \,\,0\,\,\forall \,x$ 

nên Bmax = -1 đạt được khi  $x\, = \,\frac{3}{2}\,\,;\,\,y = \,\, - \frac{1}{2}$

$B\, = \,2{\rm{x}}y\, - \,4y\, + \;16{\rm{x}}\, - \,5{{\rm{x}}^2}\, - \,{y^2}\, - \,14$

$\begin{array}{l}
a){\left( {x - 3} \right)^3} + \left( {x - 4} \right)\left( {x - 2} \right) - {\left( {3 - x} \right)^2}\\
 = {\left( {x - 3} \right)^3} + \left( {x - 4} \right)\left( {x - 2} \right) - {\left( {x - 3} \right)^2}\\
 = {\left( {x - 3} \right)^2}\left( {x - 3 - 1} \right) + \left( {x - 4} \right)\left( {x - 2} \right)\\
 = {\left( {x - 3} \right)^2}\left( {x - 4} \right) + \left( {x - 4} \right)\left( {x - 2} \right)\\
 = \left( {x - 4} \right)\left( {{x^2} - 6x + 9 + x - 2} \right)\\
 = \left( {x - 4} \right)\left( {{x^2} - 5x + 7} \right)
\end{array}$

$\begin{array}{l}
b)\left( {2a - 3b} \right)\left( {4a - b} \right) - \left( {{a^2} - {b^2}} \right) - {\left( {3b - 2a} \right)^2}\\
 = \left( {2a - 3b} \right)\left( {4a - b} \right) - \left( {{a^2} - {b^2}} \right) - {\left( {2a - 3b} \right)^2}\\
 = \left( {2a - 3b} \right)\left( {4a - b - 2a + 3b} \right) - \left( {a - b} \right)\left( {a + b} \right)\\
 = \left( {2a - 3b} \right)\left( {2a + 2b} \right) - \left( {a - b} \right)\left( {a + b} \right)\\
 = \left( {a + b} \right)\left( {4a - 6b - a + b} \right)\\
 = \left( {a + b} \right)\left( {3a - 5b} \right)
\end{array}$

$\begin{array}{l}
c){\rm{ }}{{\rm{a}}^{\rm{8}}}{\rm{ - 1}}\\
 = {\left( {{a^4}} \right)^2} - 1\\
 = \left( {{a^4} - 1} \right)\left( {{a^4} + 1} \right)\\
 = \left( {{a^2} - 1} \right)\left( {{a^2} + 1} \right)\left( {{a^4} + 1} \right)\\
 = \left( {a - 1} \right)\left( {a + 1} \right)\left( {{a^2} + 1} \right)\left( {{a^4} + 1} \right)
\end{array}$

$\begin{array}{l}
d){\rm{ }}{(x - y)^2} + 4(x - y) - 12\\
 = {(x - y)^2} + 4(x - y) + 4 - 16\\
 = {(x - y + 2)^2} - 16\\
 = (x - y + 2 + 4)(x - y + 2 - 4)\\
 = (x - y + 6)(x - y - 2)
\end{array}$

$\begin{array}{l}
e){\rm{ }}{x^2} + {y^2} + 3x - 3y - 2xy - 10\\
 = ({x^2} - 2xy + {y^2}) + (3x - 3y) - 10\\
 = {(x - y)^2} + 3(x - y) - 10\\
 = {(x - y + \frac{3}{2})^2} - \frac{{49}}{4}\\
 = (x - y + \frac{3}{2} + \frac{7}{2})(x - y + \frac{3}{2} - \frac{7}{2})\\
 = (x - y + 5)(x - y - 2)
\end{array}$

$\begin{array}{l}
\,f){\rm{ }}{x^2} - 6x - 16\,\\
 = {(x - 3)^2} - 25
\end{array}$

$\begin{array}{l}
 = (x - 3 + 5)(x - 3 - 5)\\
 = (x + 2)(x - 8)
\end{array}$

g) $A\,\, = \,\,(x + 2)(x + 3)(x + 4)(x + 5) - 24$

$\begin{array}{l}
{\rm{ = }}\,\,{\rm{[}}(x + 2)(x + 5){\rm{]}}.\,{\rm{[}}(x + 3)(x + 4){\rm{]}} - 24\,\,\\
 = \,\,({x^2}\, + \,7{\rm{x}}\, + \,10)({x^2}\, + \,7x\, + \,12)\,\, - \,24
\end{array}$

Đặt  ${{x}^{2}}\,+\,7\text{x}\,+\ 10\,=\,t\,\,$

$\Rightarrow \,A\,\,=\,\,t\,(\,t\,+\ 2)\,-\,24\,\,=\,\,{{t}^{2}}\,-\,4t\,+\,6t\,\,-\,24$          

$=\,\,t\,(\,t\,-\,4)\,+\,\,6(t\,\,-\,\,4)\,\,=\,\,(t\,\,-\,\,4)(t\,\,+\ \,6)$

$\Rightarrow $ A$=\,\,({{x}^{2}}\,+\,7\text{x}\,+\ 10\,\,-\,\,4)({{x}^{2}}\,+\,7\text{x}\,+\ 10\,\,\,+\,6)\,$

Vậy  $(x+2)(x+3)(x+4)(x+5)-24$     

$\,=\,\,({{x}^{2}}\,+\,7\text{x}\,+\ 6)({{x}^{2}}\,+\,7\text{x}\,+\ 16)$

$B\,\, = \,\,({x^2} + 6x + 5)({x^2} + 10x + 21) + 15$

$B\,\, = \,\,({x^2} + 6x + 5)({x^2} + 10x + 21) + 15$

$ = \,\,({x^2}\,\, + \,8{\rm{x}}\,\, + \,15)({x^2}\,\, + \,8{\rm{x}}\,\, + \,\,7)\,\, + \,\,15$

Đặt  ${x^2}\,\, + \,8{\rm{x}}\,\, + \,\,7\, = \,\,t$

$\begin{array}{l}
 \Rightarrow B\,\, = \,\,\,(t\,\, + \,\,8)\,t\,\, + \,\,15\,\, = \,\,\,{t^2}\,\, + \,\,8t\, + \,\,15\,\,\,\\
 = \,\,{t^2}\,\, + \,\,3t\, + \,\,5t\,\, + \,\,15
\end{array}$

$ = \,\,t\,\,(t\,\, + \,\,3)\,\, + \,\,5\,(t\,\, + \,\,3)\,\, = \,\,(t\,\, + \,\,3)(\,t\,\, + \,\,5)$

$\begin{array}{l}
 \Rightarrow B = \,\,({x^2}\,\, + \,\,8{\rm{x}}\,\, + \;\,7\,\, + \,\,3)\,({x^2}\,\, + \,\,8{\rm{x}}\,\,\, + \,\,7\,\, + \,\,5\,)\,\\
 = \,\,({x^2}\,\, + \,\,8{\rm{x}}\,\,\, + \;\,10)({x^2}\,\, + \,\,8{\rm{x}}\,\, + \;12)
\end{array}$

Vậy $({x^2} + 6x + 5)({x^2} + 10x + 21) + 15$

$ = \,\,({x^2}\,\, + \,\,8{\rm{x}}\,\,\, + \;\,10)({x^2}\,\, + \,\,8{\rm{x}}\,\, + \;12)$