a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

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

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

a,

\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b,

\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)

a: ĐKXĐ: \(\frac{3x+1}{x-1}\ge0\)

=>x>1 hoặc x<=-1/3

Đặt \(t=\sqrt{\frac{3x+1}{x-1}}\)

=>\(\frac{1}{t}=\sqrt{\frac{x-1}{3x+1}}\)

Phương trình ban đầu trở thành: \(t+6\cdot\frac{1}{t}=5\)

=>\(\frac{t^2+6}{t}=5\)

=>\(t^2+6=5t\)

=>\(t^2-5t+6=0\)

=>(t-2)(t-3)=0

=>t=2 hoặc t=3

t=2

=>\(\sqrt{\frac{3x+1}{x-1}}=2\)

=>\(\frac{3x+1}{x-1}=4\)

=>4(x-1)=3x+1

=>4x-4=3x+1

=>x=5(nhận)

t=3

=>\(\sqrt{\frac{3x+1}{x-1}}=3\)

=>\(\frac{3x+1}{x-1}=9\)

=>9(x-1)=3x+1

=>9x-9=3x+1

=>6x=10

=>x=5/3(nhận)

b: ĐKXĐ: x>=2

Đặt \(t = \sqrt{x-2} + \sqrt{x+1}\) (Điều kiện: t>0)

=>\(t^2=x-2+x+1+2\cdot\sqrt{\left(x-2\right)\left(x+1\right)}=2x-1+2\cdot\sqrt{x^2-x-2}\)

=>\(2x-7+2\sqrt{x^2 - x - 2}=t^2-6\)

Phương trình ban đầu trở thành: \(t^2-6=t\)

=>\(t^2-t-6=0\)

=>(t-3)(t+2)=0

=>t-3=0

=>t=3

=>\(\sqrt{x-2} + \sqrt{x+1} = 3\)

=>\(2x - 1 + 2\sqrt{x^2 - x - 2} = 9\)

=>\(2\sqrt{x^2 - x - 2}=10-2x\)

=>\(\sqrt{x^2 - x - 2}=5-x\)

=>\(\begin{cases}5-x\ge0\\ x^2-x-2=\left(5-x\right)^2=x^2-10x+25\end{cases}\Rightarrow\begin{cases}x\le5\\ 9x=25+2=27\end{cases}\)

=>x=3

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

$=x^2(x-5)-(x-5)=0$

$=(x-5)(x^2-1)=0$

$=(x-5)(x-1)(x+1)=0$

$\Rightarrow x=5,\ x=1,\ x=-1$

$3x^4-9x^3+9x^2-27x=0$

$3x(x^3-3x^2+3x-9)=0$

$3x[x^2(x-3)+3(x-3)]=0$

$3x(x-3)(x^2+3)=0$

Vì $x^2+3>0$ nên:

$\Rightarrow x=0$ hoặc $x=3$

b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

=>-6x+16=0

=>-6x=-16

hay x=8/3(nhận)

c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)

\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)

\(\Leftrightarrow2x^2+4x-2x^2+2=0\)

=>4x+2=0

hay x=-1/2(nhận)

a) xem lại đề

b) 3^x-1=27

=>3^x-1=3³

=>x-1=3

=>x=3+1

=>x=4

c)3^x+1=9

=>3^x+1=3²

=>x+1=2

=>x=2-1

=>x=1

a) x²=x³

⇒ x²-x³=0

⇒ x²-x².x=0

⇒ x².(x+1)=0

⇒\(\left\{{}\begin{matrix}\text{x}^2=0\\\text{x}+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\text{x}=0\\\text{x}=0+1=1\end{matrix}\right.\)

⇒ \(\text{x}\in\left\{0;1\right\}\)

b) 3^x-1=27

⇒ 3^x-1=3³

⇒ x-1=3

⇒ x= 3+1=4

c) 3^x+1=9

⇒ 3^x+1= 3³

⇒ x+1=3

⇒ x=3-1=2

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

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

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

Ta có:

$(x+4)^2-x^2(x+12)=16$

$x^2+8x+16-x^3-12x^2=16$

$-x^3-11x^2+8x=0$

$-x(x^2+11x-8)=0$

Suy ra:

$x=0$ hoặc $x^2+11x-8=0$

Giải phương trình bậc hai:

$x=\dfrac{-11\pm\sqrt{121+32}}{2}$

$=\dfrac{-11\pm\sqrt{153}}{2}$

$=\dfrac{-11\pm3\sqrt{17}}{2}$

Vậy: $x=0,\quad x=\dfrac{-11+3\sqrt{17}}2,\quad x=\dfrac{-11-3\sqrt{17}}2$

Gọi T = (x – 1)(x – 2)(x + 4)(x + 5) – 27

          = [(x – 1)(x + 4)].[(x – 2)(x + 5)] – 27

          = ( x 2 + 3x – 4).( x 2 + 3x – 10) – 27

Đặt x 2 + 3x – 7 = t   

Từ đó ta có T = (t – 3)(t + 3) – 27 = t 2 – 9 – 27 = t 2  – 36 = (t – 6)(t + 6)

Thay t = x 2 + 3x – 7 ta được

T = ( x 2 + 3x – 7 – 6)( x 2 + 3x – 7 + 6)

= ( x 2 + 3x – 13)( x 2 + 3x – 1) suy ra a = -13; b = -1 => a + b = -14

Đáp án cần chọn là: D

6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)

\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)

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

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

8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)

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

9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)

10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)

\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)

11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)

12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)

13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)