1Tìm x biết: A) x(x-2)+x-2=0 B)3x(x-1)+x-1=0 C) 5x(x-3)-x+3=0 D)2(x-3)-x^2-3x=0 2làm tính chia A) (6x^4y^3-15x^3y^2+9x^2y^2)÷3xy B) (x^3+4x^2-x-4)÷(x+

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1Tìm x biết:
A) x(x-2)+x-2=0
B)3x(x-1)+x-1=0
C) 5x(x-3)-x+3=0
D)2(x-3)-x^2-3x=0
2làm tính chia
A) (6x^4y^3-15x^3y^2+9x^2y^2)÷3xy
B) (x^3+4x^2-x-4)÷(x+1)

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Ayla 1 tháng 2021-08-12T02:16:09+00:00 1 Answers 2 views 0

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    0
    2021-08-12T02:18:03+00:00

    Đáp án:a/ $$[_{x = {\text{}} – 1}^{x = 2}$$

    b/x=3 hoặc x=1/5

     2b/ $$(x + 4)(x – 1)$$

    Giải thích các bước giải:a/ $$x\left( {x – 2} \right) + x – 2 = 0 <  =  > \left( {x – 2} \right)\left( {x + 1} \right) = 0 <  =  > [_{x + 1 = 0}^{x – 2 = 0} <  =  > [_{x = {\text{}} – 1}^{x = 2}$$

    c/$$\eqalign{
      & 5x\left( {x – 3} \right) – x + 3 = 0  \cr 
      &  <  =  > 5x\left( {x – 3} \right) – \left( {x – 3} \right) = 0  \cr 
      &  <  =  > (x – 3)(5x – 1) = 0  \cr 
      &  <  =  > [_{5x – 1 = 0}^{x – 3 = 0} <  =  > [_{x = \frac{1}{5}}^{x = 3} \cr} $$

    2b/Đk: x$\neq$ -1

    $$\frac{{{x^3} + 4{x^2} – x – 4}}{{x + 1}} = \frac{{{x^2}(x + 4) – (x + 4)}}{{x + 1}} = \frac{{(x + 4)({x^2} – 1)}}{{x + 1}} = \frac{{(x + 4)(x + 1)(x – 1)}}{{(x + 1)}} = (x + 4)(x – 1)$$

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