giải dùm mik vs ạ Bài 1: Tìm x, biết (10x^2+35x^2-2x-7).(-10x^2+18x-15x-27)=0 Bài 2: Thực hiện phép tính: a, 7x.(x^2-x+1)-(x-1).(7x+1) b, (x-1).(x-2).

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1. 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$

    

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