(x-1)(x+2)(x+3)(x+6) 
= [(x-1)(x+6)].[(x+2)(x+3)] 
=(x^2+5x-6)(x^2+5x+6) 
=(x^2+5x)^2 -6^2 = (x^2+5x)^2 -36 
vì (x^2+5x)^2 > hoặc bằng 0 => (x-1)(x+2)(x+3)(x+6) > hoặc bằng -36. 
Dấu bằng xảy ra khi (x^2+5x)^2=0 <=> x=0 hoặc x= -5

c) \(x^2-9=2\cdot\left(x+3\right)^2\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

d) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)

\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

a) \(x^2-9=2\left(x+3\right)^2\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)

\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

c) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

câu a giống Võ Đoan Nhi

câu b: 

( x² + 2x -11 ) : ( x + 2)

=> x²  + 2x -11 : ( x + 2)

=> x(x+2) -11 : ( x + 2)

Vì x( x + 2) : ( x + 2) nên  -11 : ( x + 2)

=> x + 2 thuộc ước của -11

ta lập bảng..............

\(3x+4⋮x-3\)

\(\Leftrightarrow3\left(x-3\right)+10\)\(⋮x-3\)

-Mà: \(3\left(x-3\right)⋮x-3\Rightarrow10⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(10\right)\Leftrightarrow x-3\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

-Lập bảng:.....

a, 3 - 2 | 5x - 4 | = -11

2|5x - 4| = 14

|5x - 4| = 7

Th1: 5x -4 =7

5x = 11

x= 11/5

Th2:

5x -4 =-7

5x = -3

x= -3/5

a) => 2/5x-4/=14

   =>   /5x-4/=7

  => 5x-4=7 hoac 5x-4=-7

       x=11/5         x=-3/5

bài b xét tính chẵn lẻ đi là ra liền à

ta có: 2(x-3) - 3(1-2x)=4+4(1-x)

=>    2x-6 - 3+6x = 4+4 - 4x

=>    2x+6x+4x= 4 + 4 +6+3

=>    12x         = 17

          x= 17/12

chúc bạn học tốt nhé!

a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)

b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)

c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)

\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)

Vì -50<-30<4<5

nên A<D<B<C

a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)

b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)

c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)

\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)

Vì -50<-30<4<5

nên A<D<B<C