Đáp án:
Bài 1 :
$(10x^2 +35x^2 -2x -7).(-10x^2 +18x-15x-27) =0$
$⇔(45x^2 -2x -7).(-10x^2 +3x -27) =0$
$⇔-10x^2+3x-27=0$
$⇔ -(10x^2-3x+27)=0$
$⇔-[(√10x)^2 - 2 .√10x . \dfrac{3\sqrt[]{10}}{20} + \dfrac{9}{40} +\dfrac{1071}{40}] =0$
$⇔-(√10x -\dfrac{3\sqrt[]{10}}{20})^2 - \dfrac{1071}{40} =0$
$⇔-(√10x-\dfrac{3\sqrt[]{10}}{20})^2 ≤ 0 $
$⇔-(√10x-\dfrac{3\sqrt[]{10}}{20})^2 - \dfrac{1071}{40} < 0$(loại)
$⇔ 45x^2-2x-7=0$
$⇔(√45x)² - 2 . √45x.\dfrac{\sqrt[]{5}}{15} + \dfrac{1}{45} -\dfrac{316}{45} =0$
$⇔(√45x-\dfrac{\sqrt[]{5}}{15})^2 - \dfrac{316}{45} =0$
$⇔(√45x -\dfrac{\sqrt[]{5}}{15} - \sqrt[]{\dfrac{316}{45}}).(√45x-\dfrac{\sqrt[]{5}}{15} + \sqrt[]{\dfrac{316}{45}})=0$
⇔\(\left[ \begin{array}{l}√45x -\dfrac{\sqrt[]{5}}{15} - \sqrt[]{\dfrac{316}{45}}=0\\√45x-\dfrac{\sqrt[]{5}}{15} + \sqrt[]{\dfrac{316}{45}}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{1+2\sqrt[]{79}}{45}\\x=\dfrac{1-2\sqrt[]{79}}{45}\end{array} \right.\)
Vậy $ x ∈ { \dfrac{1+2\sqrt[]{79}}{45} ; \dfrac{1-2\sqrt[]{79}}{45}}$
Bài 2 :
$a)7x.(x^2-x+1)-(x-1).(7x+1)$
$ = 7x^3 -7x^2 +7x -7x^2-x+7x+1$
$ = 7x^3 +13x+1$
$b) (x-1)(x-2).(x-3)$
$ = [(x-1).(x-2)](x-3)$
$ = [x^2-2x-x+2].(x-3)$
$ = (x^2-3x+2).(x-3)$
$ = x^3 -3x^2-3x^2+9x+2x-6$
$ = x^3 -6x^2+11x-6$
