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

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

`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

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

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

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

giúp mink vs ạ 

Em vào phần \(\Sigma\) ở góc bên trái để đăng đề bài cho dễ nhìn nhé

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 : 

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

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

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

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

\(\left(+\right)A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right).\)

\(A=8x^3-6x^2-18x+27-8x^3+2\)

\(A=6x^2-18x+29\)

\(\left(+\right)B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x-1\right)\left(x+1\right)\)

\(B=x^3-3x^2+3x+1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)

\(B=-6x^2+6x^2-6\)

\(B=-6\)

Ta có: \(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)\)

\(=6x^2-2x-\left(2x^2-6x+x-3\right)\)

\(=6x^2-2x-2x^2+5x+3\)

\(=4x^2+3x+3\)

Ta có: \(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)\)

\(=3x^2-6x-\left(4x^2-4x+2x-2\right)\)

\(=3x^2-6x-4x^2+2x+2\)

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

\(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)=6x^2-2x-2x^2+5x+3=4x^2+3x+3\)

\(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)=3x^2-6x-4x^2+2x-2=-x^2-4x-2\)

cậu xem câu hỏi của Quỳnh Phạm nha

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

\(=3x-y-y-x+2x^2-2x\)

\(=2x^2-2y\)