1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)

A) -2x(3x+2)(3x-2)+5(x+2)² - (x-1)(2x+1)(2x+1)

= -2x(9x²-4)+5(x²+4x+4) - (x-1)(4x²-1)

= -18x³+8x+5x²+20x+20-(4x³-x-4x²+1)

= -18x³+5x²+28x+20-4x³+x+4x²+1

= -22x³+9x²+29x+21

B) (7x-8)(7x+8)-10(2x+3)²+5x(3x-2)²-4x(x-5)²

= 49x² - 64 -10(4x²+ 12x + 3) + 5x(9x² - 12x +4) - 4x(x² - 10x +25)

= 49x² - 64 -40x² - 120x - 30 + 45x³ - 60x² - 20x - 4x³ + 40x² -100x

= 41x³ -11x² -240x -94

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

\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)

\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)

\(-5x^4-x^3+5x^2+20x-9\)

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

\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)

\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)

\(-40x^4+36x^3+82x^2+6x-11\)

a) Đặt A(x)=0

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=-\dfrac{5}{4}\)

b) Đặt B(x)=0

\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)

\(\Leftrightarrow6x-3-2x-2=0\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

Yêu cầu đề là gì vậy bạn?

dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

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

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)