Phiếu bài tập tuần Toán 8 – Tuần 07

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

b) $left$2a-3bright$left$4a-bright$-left$a^2-b^{2}right$-{{left$3b-2aright$}^{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$x^2,–,3rmx),rm+,rm9,,rm=,,,rm–2left(x^2,–,2.,x.,frac32,+,frac94right$,, + ,frac{9}{2}, + ,9\
 = , – 2{left$x,–,frac32right$^2}, + ,frac{{27}}{2}, le ,,frac{{27}}{{2,}},,,forall ,x
end{array}$

Vì $ – 2{left$x,–,frac32right$^2}, le ,0,$ nên  $A,, le ,,frac{{27}}{2}$

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

$begin{array}{l}
B, = ,$–{rmx}^2,+;2rmxy,–,y^2$, + 4$x,–,y$,, + ,12x, – ,4{x^2}, – ,14\
B,, = ,, – $$({rmx}^2,–;2rmxy,+,y^2),–4(x,–,y),+,4$$, – $4x^2,–,12x+9$,, – ,,1
end{array}$

$B, = , – [{$x,–,y$^2}, – ,2.$x–,y$.2, + ,{2^2}{rm{]}}, – ,{$2rmx,–3$^2}, – ,1$

$B,, = ,, – {$x,–,y,–,2$^2}, – ,{$2rmx,–,3$^2}, – ,1$

Vì $ – {$x,–,y,–,2$^2}, le ,0,,, – {$2rmx,–,3$^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–3right$^3} + left$x–4right$left$x–2right$ – {left$3–xright$^2}\
 = {left$x–3right$^3} + left$x–4right$left$x–2right$ – {left$x–3right$^2}\
 = {left$x–3right$^2}left$x–3–1right$ + left$x–4right$left$x–2right$\
 = {left$x–3right$^2}left$x–4right$ + left$x–4right$left$x–2right$\
 = left$x–4right$left$x^2–6x+9+x–2right$\
 = left$x–4right$left$x^2–5x+7right$
end{array}$

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

$begin{array}{l}
c){rm{ }}{{rm{a}}^{rm{8}}}{rm{ – 1}}\
 = {left$a^{4}right$^2} – 1\
 = left$a^4–1right$left$a^4+1right$\
 = left$a^2–1right$left$a^2+1right$left$a^4+1right$\
 = left$a–1right$left$a+1right$left$a^2+1right$left$a^4+1right$
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\
 = {$x–y+frac32$^2} – frac{{49}}{4}\
 = $x–y+frac32+frac72$$x–y+frac32–frac72$\
 = $x–y+5$$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,+,7rmx,+,10$$x^2,+,7x,+,12$,, – ,24
end{array}$

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

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

$=,,t,$,t,-,4$,+,,6$t,,-,,4$,,=,,$t,,-,,4$$t,,+,6$$

$Rightarrow $ A$=,,$x^2,+,7textx,+10,,-,,4$$x^2,+,7textx,+10,,,+,6$,$

Vậy  $$x+2$$x+3$$x+4$$x+5$-24$     

$,=,,$x^2,+,7textx,+6$$x^2,+,7textx,+16$$

$B,, = ,,$x^2+6x+5$$x^2+10x+21$ + 15$

$B,, = ,,$x^2+6x+5$$x^2+10x+21$ + 15$

$ = ,,$x^2,,+,8rmx,,+,15$$x^2,,+,8rmx,,+,,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,,+,,8rmx,,+;,7,,+,,3$,$x^2,,+,,8rmx,,,+,,7,,+,,5,$,\
 = ,,$x^2,,+,,8rmx,,,+;,10$$x^2,,+,,8rmx,,+;12$
end{array}$

Vậy $$x^2+6x+5$$x^2+10x+21$ + 15$

$ = ,,$x^2,,+,,8rmx,,,+;,10$$x^2,,+,,8rmx,,+;12$$