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