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

\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)

\(\Rightarrow A=x^3+8-x^3+2\)

\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)

\(\Rightarrow A=10\)

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

\(=x^3+8-x^3+2\)

\(=10\)

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

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

\(=\left(x^3+8\right)\left(x^3-8\right)\)

\(=x^6-64\)

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

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

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

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

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

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

\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)

\(=-9x^2\)

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

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

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

\(=-4x^2\)

1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)

\(=-x^2+2x+5x-10+x^2-49=7x-59\)

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

\(=9x^2+6x+1-9x^2+4=6x+5\)

=>7x-59=6x+5

=>x=64

2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)

\(=5x^2+5x-x-1-2x^2+12x-9\)

\(=3x^2+16x-10\)

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

\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)

\(=3x^2-4x-18\)

=>16x-10=-4x-18

=>20x=-8

hay x=-2/5

Cảm ơn nhiều nha :))

a) ta có:

\(x^2y^2+1-x^2-y^2\)

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

\(=\left(y+1\right)\left(y-1\right)\left(x-1\right)\left(x+1\right)\)

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

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

c) tương tự b

\(2x^2-4x=2x\left(x-2\right)\)

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

\(10\left(x-y\right)-6x\left(y-x\right)=10\left(x-y\right)+6x\left(x-y\right)=\left(10+6x\right)\left(x-y\right)=2\left(x-y\right)\left(3x+5\right)\)\(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

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

\(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

\(x^2-7x-y^2+7y=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\)

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

thanks

\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)

\(=\frac{-2}{x^2}\)

\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)

\(=x\left(x-3\right)\)

\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

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

\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x+3}{x+1}\)

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

******************************************************

a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)

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

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

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

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

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

c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)

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

d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)

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

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

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

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

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

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

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

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

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

gắp cái gì

Bài 1. Rút gọn:

\(a, x\left(1-x\right)+6\left(x+3\right)\left(x+3\right)\)

\(=x-x^2+6\left(x^2+6x+9\right)\)

\(=x-x^2+6x^2+36x+54\)

\(=5x^2+37x+54\)

\(b, \left(2-3x\right)\left(2+3x\right)-\left(x+5\right)\left(x-5\right)\)

\(=\left(4-9x^2\right)-\left(x^2-25\right)\)

\(=-10x^2+29\)

\(c, \left(3x+1\right)\left(x+5\right)-\left(x-1\right)\left(x+1\right)\)

\(=3x^2+15x+x+5-x^2+1\)

\(=2x^2+16x+6\)

\(d,\left(2-3x\right)\left(2x+3\right)+6\left(x-1\right)^2\)

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

\(=4x+6-6x^2-9x+6x^2-12x+6\)

\(=-17x+12\)

\(e, x\left(5-x\right)-\left(2x+2\right)\left(3x+2\right)-\left(x-2\right)\left(x+2\right)\)

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

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

\(=-8x^2-5x\)

Bài 2: 

a: VT\(=x^3-xy+x^2y^2-y^3-x^3+y^3-x^2y^2\)

=-xy

b: \(VT=x^2+6xy+9y^2-x^2+9y^2-6xy=18y^2=VP\)