Xét x=1,x=2,1<x<2,x<1,x>2 

cam on ban nha. ban hoc truong nao vay?

a) [x(x+1].[(x-1)(x+2)]=24

(x²+x)(x²+x+2)=24

Dat x²+x=a , ta dc: a(a+2)=24

=> a²+2a-24=0

=> (a-4)(a+6)=0

=> a=4 hoac a=-6

Thay vao roi tu tim x nha

b)

thanks

Nếu là bài tìm x thì mình xin làm như sau

a) Ta có: \(x^2+4x+4=6\left(x+2\right)\)

\(\Rightarrow\left(x+2\right)^2=6\left(x+2\right)\)

\(\Rightarrow\left(x+2\right)^2-6\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+2-6\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;4\right\}\)

b) ta có: \(27^3-72x=0\)

\(\Rightarrow19683-72x=0\)

hay \(72x=19683\)

hay x=\(\frac{19683}{72}=273,375\)

Vậy: \(x=273,375\)

có bao nhiêu số hạng vậy bạn

Lời giải:

\((x^3-x^2)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2(x-1)-4(x^2-2x+1)=0\)

\(\Leftrightarrow x^2(x-1)-4(x-1)^2=0\)

\(\Leftrightarrow (x-1)[x^2-4(x-1)]=0\)

\(\Leftrightarrow (x-1)(x^2-4x+4)=0\)

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

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

a, (1-x)(5x+3)= (3x-8)(1-x)

<=> (1-x) (5x+3) - (3x-8)(1-x) =0 <=> (1-x) (2x+11) = 0

\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)

Vậy.........

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

<=> 3x = 24<=> x=8

Vậy .......

c,x^2+ x^3+x+1=0

<=> x^2 (x+1) +(x+1) =0 <=> (x^2 +1)(x+1) =0

<=> x+1 =0 => x=-1

Vậy.......

d, \(\dfrac{x-3}{x+3}-\dfrac{2}{x-3}=\dfrac{3x+1}{9-x^2}\)

\(\Leftrightarrow x^2-6x+9-2x-6=-3x-1\)

\(\Leftrightarrow x^2-5x+4=0\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

Vậy...........