a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

=>-x^2(x-2)-(x-2)(x+2)=0

=>(x-2)(-x^2-x-2)=0

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

$\Leftrightarrow (2x+1-x)(2x+1+x)=0$

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

$\Leftrightarrow (x-1)^2-2^2=0$

$\Leftrightarrow (x-1-2)(x-1+2)=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

$\Leftrightarrow (x^2-2x)-(3x-6)=0$

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

Bài 2:

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

  Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

  Vậy \(A_{Min}=-2\) khi \(x=-1\)

Bài 1:

a) Ta có: \(2x^2-6=0\)

\(\Leftrightarrow2x^2=6\)

\(\Leftrightarrow x^2=3\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)

chỗ dấu suy ra thứ 2 e ko hiểu lắm ạ 

a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)

=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)

=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22

=>-x^2+59x+14-8x^2+5x+22=0

=>-9x^2+54x+36=0

=>x^2-6x-4=0

=>\(x=3\pm\sqrt{13}\)

b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)

=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)

=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32

=>x^2+6x+19=x^2+4x-32

=>2x=-51

=>x=-51/2

kh hiểu bn ơi

vậy mik đăng lại

lớp 8 có pt bậc 2 ak??

Có nhưng giải bằng PT tích nhé

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)