a: \(\left(x^2-2x+1\right)\left(x^2+bx+c\right)\)

\(=x^4+bx^3+cx^2-2x^3-2b\cdot x^2-2x\cdot c+x^2+bx+c\)

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

Theo đề, ta có: b-2=-2; c-2b+1=2; b-2=-2; a=c

=>b=0; c=1; a=c=1

b: (x-2)(x^2+bx+c)+a

\(=x^3+bx^2+cx-2x^2-2bx-2c+a\)

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

Theo đề ta có: b-2=3; c-2b=-1; -2c+a=-3

=>b=5; c=-1+2b=-1+10=9; a=-3+2c=-3+2*9=15

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

\(=x^4+bx^3+cx^2-2x^3-2bx^2-2cx+x^2+bx+c\)

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

Đồng nhất phần hệ số ta được : \(b-2=-2;c-2b+1=2;-2c+b=-2;c=a\)

\(\Rightarrow b=0;c=1;a=1\)

`a)3x(2x^2-3x+4)`

`=6x^3-9x^2+12x`

______________________________________________

`b)(x+3)^2+(3x-2)(x+4)`

`=x^2+6x+9+3x^2+12x-2x-8`

`=4x^2+16x+1`

______________________________________________

`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`

`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`

`=[2x^2-2]/[x^2-1]`

`=2`

hếp

a/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\left(2x\right)^2.\left(4x\right)^3}{x^4}}{\dfrac{\left(3x\right)^2\left(5x^2\right)}{x^4}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{4^4.x}{45}=\pm\infty\)

b/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{2x^2}{x^3}+\dfrac{x}{x^3}}}{\dfrac{2x}{x}-\dfrac{2}{x}}=\dfrac{1}{2}\)

c/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\sqrt[3]{\left(x^3+2x^2\right)^2}}{x^2}+\dfrac{x\sqrt[3]{x^3+2x^2}}{x^2}+\dfrac{x^2}{x^2}}{\dfrac{3x^2}{x^2}-\dfrac{2x}{x^2}}=\dfrac{1+1+1}{3}=1\)

d/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(-3x\right)^3x^2}{x^5}}{-\dfrac{4x^5}{x^5}}=\dfrac{-27}{-4}=\dfrac{27}{4}\)

e/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(2x\right)^{20}.\left(3x\right)^{20}}{x^{50}}}{\dfrac{\left(2x\right)^{50}}{x^{50}}}=0\)

g/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{8x^3.\left(4x^5\right)^9}{x^{47}}}{\dfrac{11x^{47}}{x^{47}}}=+\infty\)

\(a.\Leftrightarrow x^2+x-6+2x^2+4x+2=x^2-6x+9-2x^2+4x\)

\(\Leftrightarrow4x^2+7x-13=0\)(pt vô nghiệm)

\(b.\Leftrightarrow x^3+3x^2+3x+1-x^2+2x+8=x^3-8+2x^2\)

\(\Leftrightarrow5x=-17\Rightarrow x=\frac{-17}{5}\)

Đặt \(t=x^2+2x+2\left(t\ge1\right)\)

\(c.\Leftrightarrow\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\)\(\Leftrightarrow\frac{t^2-1+t^2}{t^2+t}=\frac{7}{6}\)\(\Leftrightarrow12t^2-6=7t^2+7t\)

\(\Leftrightarrow5t^2-7t-6=0\Rightarrow\orbr{\begin{cases}t=2\left(tm\right)\\t=\frac{-3}{5}\left(l\right)\end{cases}}\)

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

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

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

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

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

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

\(=x^3+6x^2+7x-6\)

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

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

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)