\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

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

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

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

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)

a) x Î{0;3}.

b) xÎ{0;-9}.

c) x Î{-l; 11}.

d) x = 13.

a) 4x(x+1)=8(x+1)

<=>4x(x+1)-8(x+1)=0

<=>(4x-8)(x+1)=0

<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)

Vậy...

b)x(x-1)-2(1-x)=0

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

<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)

Vậy...

c)5x(x-2)-(2-x)=0

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

<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)

d)5x(x-200)-x+200=0

<=>(5x-1)(x-200)=0

<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)

e)\(x^3+4x=0 \)

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

\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)

Vậy x=0

f)\((x+1)=(x+1)^2\)

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

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

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

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

Vậy....

a) -3(x-4)+5(x-1)=-7

=>-3x+12+5x-5=-7

=>2x+7=-7

=>2x=-14=>x=-7

b) -4./x-8/+12=0

=>/x-8/=3

=>x-8=3 hoặc -3

(tự tính) 

a, \(\left(-7\right)\left(5-x\right)< 0\)

\(< =>5-x>0< =>x< 5\)

b, \(11⋮x-1< =>x-1\inƯ\left(11\right)\in\left\{-11;-1;1;11\right\}\) ( \(x\ne1\) )

\(x\in\left\{-10;0;2;12\right\}\)

c, \(x+8⋮x+1< =>x+1+7⋮x+1\)

\(< =>7⋮x+1< =>x+1\inƯ\left(7\right)\in\left\{-7;-1;1;7\right\}\left(x\ne-1\right)\)

\(< =>x\in\left\{-8;-2;0;6\right\}\)

d, \(\left(x+2\right)\left(5-x\right)>0\)

Chưa học lập bảng xét dấu nên xét TH em nhé !

Nhận thấy ( x + 2 ) ( 5 - x ) > 0 nên x + 2 và 5 - x phải cùng dấu

TH1 : \(\left\{{}\begin{matrix}x+2>0\\5-x>0\end{matrix}\right.< =>\left\{{}\begin{matrix}x>-2\\x< 5\end{matrix}\right.< =>-2< x< 5}\)

TH2:

\(\left\{{}\begin{matrix}x+2< 0\\5-x< 0\end{matrix}\right.< =>\left\{{}\begin{matrix}x< -2\\x>5\end{matrix}\right.< =>x\in\varnothing\)

Từ 2 TH ta kết luận { x | -2 < x < 5 }

Điều kiện về x là gì bạn? Số nguyên, số tự nhiên, số hữu tỉ,...?

a) \(x^3=-8\)

\(\Rightarrow x=-2\)

b) \(x^{10}-x^8=0\)

\(\Leftrightarrow x^{10}=x^8\)

\(\Rightarrow x=1\)

\(\Rightarrow x=-1\)

\(\Rightarrow x=0\)

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

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

\(thay\) x vào thì ta sẽ có giá trị thỏa mãn.

d) \(2^5-4=2^n-1\)

\(\Rightarrow32-4=2^{n-1}\)

\(\Rightarrow28=2^{n-1}\)

\(\leftrightarrow\) \(n\) không có giá trị thỏa mãn

a) x³ = -8

=> x = -2

b) x¹⁰ - x⁸ = 0

x¹⁰ = x⁸

=> x = 1 ; x = -1 ; x = 0

c) ( x + 1 )⁵ + ( x + 1 )² = 0

( x + 1 )⁵ = - ( x + 1 )²

Làm tiếp như phần b

d) 2⁵ - 4 = 2^n-1

32 - 4 = 2^n-1

28 = 2^n-1

=> n không có giá trị thỏa mãn