Mình cần gấp ạk

A và B

a: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(120-25y^2-10y\right)-\left(120-25y^2-10y\right)\)

\(=\left(120-25y^2-10y\right)\left(x^3-x^2+x-1\right)\)

\(=-\left[\left(25y^2+10y+1\right)-121\right]\left[x^2\left(x-1\right)+\left(x-1\right)\right]\)

\(=-\left(5y-10\right)\left(5y-12\right)\left(x-1\right)\left(x^2+1\right)\)

\(=-5\left(y-2\right)\left(5y-12\right)\left(x-1\right)\left(x^2+1\right)\)

b: \(x^4-14x^3+71x^2-154x+120\)

\(=x^4-5x^3-9x^3+45x^2+26x^2-130x-24x+120\)

\(=\left(x-5\right)\left(x^3-9x^2+26x-24\right)\)

\(=\left(x-5\right)\left(x^3-4x^2-5x^2+20x+6x-24\right)\)

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

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

\(\dfrac{9}{4}x^2-25y^2=\left(\dfrac{3}{2}x-5y\right)\left(\dfrac{3}{2}x+5y\right)\)

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

a)x²-4x=x(x-4)

b)x²-5xy+x-5y=x(x-5y)+(x-5y)=(x+1)(x-5y)

c)x²-10xy-49+25y²=x²-10xy+25y²-49

=(x-5)²-7²=(x-5-7)(x-5+7)

=(x-12)(x+2)

\(a,Sửa:x^2+4xy-9+4y^2=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ b,=\left(x-6y\right)^2-1=\left(x-6y-1\right)\left(x-6y+1\right)\\ c,=36-\left(x-5y\right)^2=\left(6-x+5y\right)\left(6+x-5y\right)\)

b: \(=\left(\dfrac{1}{2}x-5y\right)^2\)

\(a,=x^2-4-x^2+2x+4=2x\\ b,=\left(x-5y\right)^2:\left(5y-x\right)=\left(5y-x\right)^2:\left(5y-x\right)=5y-x\\ c,Sửa:\left(28x-9x^2+x^3-30\right):\left(x-3\right)\\ =\left(x^3-3x^2-6x^2+18x+10x-30\right):\left(x-3\right)\\ =\left(x-3\right)\left(x^2-6x+10\right)\left(x-3\right)=x^2-6x+10\)

\(x^4-x^2+10x-25\)

\(=x^4-\left(x^2-10x+25\right)\)

\(=\left(x^2\right)^2-\left(x^2-2\cdot5\cdot x+5^2\right)\)

\(=\left(x^2\right)^2-\left(x-5\right)^2\)

\(=\left[x^2-\left(x-5\right)\right]\left[x^2+\left(x-5\right)\right]\)

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

a)4x²+4x+1-x²-10x-25=0

`<=>(2x+1)^2-(x+5)^2=0`

`<=>(2x+1-x-5)(2x+1+x+5)=0`

`<=>(x-4)(3x+6)=0`

`<=>(x-4)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 
b)(x^2+x+7)(x^2+x-7)=(x²+x)²-7x

`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`

`<=>-7^2=-7x`

`<=>-49=-7x`

`<=>x=7`

Vậy x=7