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

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

\(A=5\)

=> giá trị biểu thức ko phụ thuộc vào biến x

b) \(A=x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)

=> \(A=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x\)

=> \(A=\)16

vậy giá trị của biểu thức A ko phụ thuộc vào biến x

a) x(2x+1)-x²(x+2)+(x³-x+3)= 2x²+x-x³-2x²+x³-x+3= 3

b)x (3x²-x+5)-(2x³+3x-16)-x(x²-x+2)= 3x³-x²+5x-2x³-3x+16-x³+x²-2x= 16

a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)

b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)

c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

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

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

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

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

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

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

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

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

a) \(\left(2x^3-x^2+5x\right):x\)

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

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

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

b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)

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

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

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

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

c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)

\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)

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

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

d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)

\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)

\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)

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

\(=-2x^2+4xy-6y^2\)

e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)

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

\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)

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

f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)

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

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

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