a: 3(x+7)-2x+5>0

=>3x+21-2x+5>0

=>x+26>0

=>x>-26

Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)

=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)

=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)

=>\(4x+8-9x-27< 8x-8-3x+12\)

=>-5x-19<5x+4

=>-10x<23

=>\(x>-\dfrac{23}{10}\)

b: \(3x+2+\left|x+5\right|=0\left(1\right)\)

TH1: x>=-5

(1) trở thành: 3x+2+x+5=0

=>4x+7=0

=>\(x=-\dfrac{7}{4}\left(nhận\right)\)

TH2: x<-5

=>x+5<0

=>|x+5|=-x-5

Phương trình (1) sẽ trở thành:

\(3x+2-x-5=0\)

=>2x-3=0

=>2x=3

=>\(x=\dfrac{3}{2}\)

\(\dfrac{3x+4}{7}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow\dfrac{2\left(3x+4\right)}{14}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow2\left(3x+4\right)\le5x-19\)

\(\Leftrightarrow6x+8\le5x-19\)

\(\Leftrightarrow x\le-27\)

Vậy \(S=\left\{x|x\le-27\right\}\)

\(\dfrac{3x+4}{7}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow\dfrac{6x+8}{14}\le\dfrac{5x-19}{14}\)

\(\Leftrightarrow6x+8\le5x-19\)

\(\Leftrightarrow6x-5x\le-19-8\)

\(\Leftrightarrow x\le-27\)

⇔2(3x+4)14≤5x−1914⇔2(3x+4)14≤5x−1914

$\iff 2\left(3x+4\right)\le 5x-19$

$\iff 6x+8\le 5x-19$

$\iff x\le -27$

Vậy S={x|x≤−27}

N1và N2:3 đầy;1 nửa;3 không

N3:1 đầy;5  nửa;1 không

Còn có cách này

n1 và n2 " 2 đầy, 3 nữa ,2 ko" .. n3 " 3 đầy , 1 nữa ,3 ko"

a: =>(x^2-1)(x^2-4)=0

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

=>\(x\in\left\{1;-1;2;-2\right\}\)

b: =>2x^4-4x^2+x^2-2=0

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

=>x^2-2=0

=>\(x=\pm\sqrt{2}\)

c: =>(căn x-6)(căn x+1)=0

=>căn x-6=0

=>x=36

Bài làm:

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

=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\) hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy tập nghiệm PT \(S=\left\{-2;-1;0;1\right\}\)

b) Nhận thấy \(\left(x-1\right)^4+\left(x-2\right)^4=0\)

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

Mà \(\hept{\begin{cases}\left(x-1\right)^4\ge0\\-\left(x-2\right)^4\le0\end{cases}\left(\forall x\right)}\) 

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^4=0\\-\left(x-2\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=2\end{cases}}\) (vô lý)

=> không tồn tại x thỏa mãn PT

a) x( x - 1 )( x + 1 )( x + 2 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\), \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\), \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

b) ( x - 1 )⁴ + ( x - 2 )⁴ = 0

<=> ( x - 1 )⁴ = -( x - 2 )⁴

\(\hept{\begin{cases}\left(x-1\right)^4\ge0\\-\left(x-2\right)^4\le0\end{cases}\forall}x\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=2\end{cases}}\)( mâu thuẫn )

=> Phương trình vô nghiệm

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

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=0\\x+2=0\end{cases};x=0}\Leftrightarrow\hept{\begin{cases}x=1\\x=-1\\x=-2\end{cases};x=0}\)

b, \(\left(x-1\right)^4+\left(x-2\right)^4=0\)

Ta có  : \(\left(x-1\right)^4\ge0;\left(x-2\right)^4\ge0\)

\(< =>\left(x-1\right)^4+\left(x-2\right)^4\ge0\)

Dấu = xảy ra khi \(< =>\hept{\begin{cases}x=1\\x=2\end{cases}}\)(vô lý)

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

\(2x^2-14+3x=0\)

\(2x^2+3x-14=0\)

\(\Delta=b^2-4ac=3^2-4.2.\left(-14\right)=9+112=121>0\)

Nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-3-\sqrt{121}}{2.2}=\frac{-3-11}{4}=-\frac{7}{2}\)

\(x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-3+\sqrt{121}}{2.2}=\frac{-3+11}{4}=\frac{8}{4}=2\)

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

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

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

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

\(\Leftrightarrow\left(x-2\right).7=0\)

\(\Leftrightarrow7x-14=0\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)

=> pt vô nghiệm

b) \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)

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

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

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

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(x^4-1+x^4-81=0\)

\(2x^4-82=0\)

\(2x^4=82\)

\(x^4=41\)

\(x=\sqrt[4]{41}\)

\(\Rightarrow\)vô nghiệm