`@` `\text {Ans}`

`\downarrow`

\(B=(x+1)^2-2(2x-1)(1+x)+4x^2-4x+1\)

`= x^2 + 2x + 1 - 2(2x^2 + x - 1) + 4x^2 - 4x + 1`

`= 5x^2 - 2x + 2 - 4x^2 - 2x + 2`

`= x^2 - 4x + 4`

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

\(=\left(x+1\right)^2-2\left(x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left(x+1-2x+1\right)^2\)

\(=\left(2-x\right)^2\)

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

\(=\left(x+1\right)^2-2\left(x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left(x+1-2x+1\right)^2=\left(2-x\right)^2\)

\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-2\right)\\ =\left[\left(2x\right)^2-2\cdot2x\cdot1+1^2\right]-\left(4x^2-8x+x-2\right)\\ =4x^2-4x+1-4x^2+8x-x+2\\ =3x+3\)

\(\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\right)-\left(x^3-x^2+x+x^2-x+1\right)\\ =x^3-9x^2+27x-27-x^3+x^2-x-x^2+x-1\\ =-10x^2+27x-28\)

`P= 8x^3 -4x^2 +2x+1+x^3+x^2-x+1`

`P=9x^3 -3x^2+x+2`

\(\text{ P = (2x-1).4x^2+2x+1+(x+1)x^2-x+1}\)

\(\text{P =}\) \(\text{[(2x-1) . 4x^2 ]}\)\(\text{[(x+1) .x^2]}\)

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

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

\(\text{P =}\) \(\text{11x^3 - 2x^2 + x + 2}\)

mình mới lên lớp 6

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

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

mình mới lên lớp 6 cũng thanh luôn

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

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

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

Chọn đáp án D.

\(=4x^2+4x-5x-5-\left(3x^2+x+9x+3\right)\)

\(=4x^2-x-5-3x^2-10x-3=x^2-11x-8\)

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: \(x\ne\pm1\)

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

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

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

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