KO

Để rút gọn biểu thức, ta sẽ thực hiện các phép tính và kết hợp các thành phần tương tự: P(2x-1).4x^2 + 2x + 1 + (x+1)x^2 - x + 1 = P(8x^3 - 4x^2) + 2x + 1 + x^3 + x^2 - x + 1 = P(8x^3) - P(4x^2) + x^3 + (2x-x) +(1+1) = **8Px^3 - 4Px^2**+ x^3 **+ x**+ **2** Vậy biểu thức đã được rút gọn thành: **8Px³ - 4Px²+x³+x+2**

ok bạn

ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)

\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)

https://sg.docworkspace.com/l/sIM-LioBEocfloAY

ĐKXĐ : \(x\ne\pm\frac{1}{2}\)

\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)

\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)

\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)

\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)

\(E=\frac{8x^3+1}{1+4x^2}\)

Study well 

E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)

E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)

E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)

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

Góp ý : 

Cô giáo ơi

Cô tính nhầm rồi 2x + 2x = 4x cô ạ chứ không phải 8x đâu cô

mong cô xem lại ạ 

a,sửa đề :  \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)

\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)

b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)

\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)

\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)

\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)

\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)

\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)

\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)

\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x+1}{2x}\)

b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:

\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)

(2x+1)²+(2x-1)²-2(4x²-1)

= (4x²+2x+1)+(4x²-2x+1)-8x²+2

= 4x²+2x+1+4x²-2x+1-8x²+2

= 1+1+2

= 4

(2x+1)² + (2x-1)² - 2(4x²-1)

= (2x+1)² + (2x-1)² - 2[(2x)²-1]

= (2x+1)² + (2x-1)² - 2(2x+1)(2x-1)

= (2x+1-2x+1)²

= 2²

= 4