A = (x-2/x+2 – 4x^2/x^2-4 – 2+x/x-2).x-2/2x^2-x. rút gọn bt A 06/12/2021 Bởi Gianna A = (x-2/x+2 – 4x^2/x^2-4 – 2+x/x-2).x-2/2x^2-x. rút gọn bt A
Đáp án: A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$ Giải thích các bước giải: A=( $\frac{x-2}{x+2}$ – $\frac{4²}{x²÷x^4}$ – $\frac{2+x}{x-2}$ ) ×$\frac{x-2}{2x^2-x}$ MTC : (x²-4)(x²-x^4) A= ( $\frac{(x-2)(x-2)(x^2-x^4)}{(x²-4)(x²÷x^4)}$ – $\frac{4x²(x²-4)}{(x²-4)(x²÷x^4)}$ -$\frac{(2+x)(x-2)(x^2÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A=( $\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4}{(x²-4)(x²÷x^4)}$ – $\frac{4x^4-16x^2}{(x²-4)(x²÷x^4)}$ – $\frac{(x²-4)(x²÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A= ($\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4-4x^4+16x^2-x^4+x^6+4x^2-4x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A=( $\frac{2x^5+24x^2-2x^3-12x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A=( $\frac{(24x^2-12x^4)+(2x^5-2x^3)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A= ($\frac{12x^2(2x^2-1)+2x^3(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A= ($\frac{(12x^2+2x^3)(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$ A=( $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x²÷x^4)}$) $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x^-2)}$× $\frac{x-2}{2x^(2-x)}$ A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)}$× $\frac{1}{2x^(2-x)}$ A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$ Vậy A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$ CHO MK 5* VÀ CTLHN NHÁ CHÚC BẠN HỌC TỐT Bình luận
Đáp án: $A = \dfrac{12}{(x+2)(1- 2x)}$ Giải thích các bước giải: $\begin{array}{l}A = \left(\dfrac{x-2}{x+2}-\dfrac{4x^2}{x^2 – 4}-\dfrac{2+x}{x-2}\right)\cdot\dfrac{x-2}{2x^2 -x}\qquad (x \ne 0;\, x \ne \pm 2) \\ \to A = \left[\dfrac{(x-2)^2}{(x+2)(x-2)}-\dfrac{4x^2}{(x+2)(x-2)}-\dfrac{(2+x)^2}{(x+2)(x-2)}\right]\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{x^2 – 4x + 4 – 4x -x^2 – 4x – 4}{(x+2)(x-2)}\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{-12x}{(x+2)(x-2)}\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{12}{(x+2)(1- 2x)} \end{array}$ Bình luận
Đáp án:
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
Giải thích các bước giải:
A=( $\frac{x-2}{x+2}$ – $\frac{4²}{x²÷x^4}$ – $\frac{2+x}{x-2}$ ) ×$\frac{x-2}{2x^2-x}$
MTC : (x²-4)(x²-x^4)
A= ( $\frac{(x-2)(x-2)(x^2-x^4)}{(x²-4)(x²÷x^4)}$ – $\frac{4x²(x²-4)}{(x²-4)(x²÷x^4)}$ -$\frac{(2+x)(x-2)(x^2÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4}{(x²-4)(x²÷x^4)}$ – $\frac{4x^4-16x^2}{(x²-4)(x²÷x^4)}$ – $\frac{(x²-4)(x²÷x^4)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{x^4-x^6-2x^3+2x^5+4x^2-4x^4-4x^4+16x^2-x^4+x^6+4x^2-4x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{2x^5+24x^2-2x^3-12x^4}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{(24x^2-12x^4)+(2x^5-2x^3)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{12x^2(2x^2-1)+2x^3(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A= ($\frac{(12x^2+2x^3)(2x^2-1)}{(x²-4)(x²÷x^4)}$) ×$\frac{x-2}{2x^2-x}$
A=( $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x²÷x^4)}$) $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)(x^-2)}$× $\frac{x-2}{2x^(2-x)}$
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{(x²-4)}$× $\frac{1}{2x^(2-x)}$
A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
Vậy A= $\frac{2x^2(6-x^3)(2x+1)(2x-1)}{2x^(2-x)(x²-4)}$$\frac{x-2}{2x^2-x}$
CHO MK 5* VÀ CTLHN NHÁ
CHÚC BẠN HỌC TỐT
Đáp án:
$A = \dfrac{12}{(x+2)(1- 2x)}$
Giải thích các bước giải:
$\begin{array}{l}A = \left(\dfrac{x-2}{x+2}-\dfrac{4x^2}{x^2 – 4}-\dfrac{2+x}{x-2}\right)\cdot\dfrac{x-2}{2x^2 -x}\qquad (x \ne 0;\, x \ne \pm 2) \\ \to A = \left[\dfrac{(x-2)^2}{(x+2)(x-2)}-\dfrac{4x^2}{(x+2)(x-2)}-\dfrac{(2+x)^2}{(x+2)(x-2)}\right]\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{x^2 – 4x + 4 – 4x -x^2 – 4x – 4}{(x+2)(x-2)}\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{-12x}{(x+2)(x-2)}\cdot\dfrac{x-2}{x(2x-1)}\\ \to A = \dfrac{12}{(x+2)(1- 2x)} \end{array}$