Ta có: \(\frac{1}{3^2}< \frac{1}{2.3}\)

           \(\frac{1}{4^2}< \frac{1}{3.4}\)

            .....................

            \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)

Đặt \(B=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)

           \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)

             \(=\frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)

\(\text{Ta có: }n^2>n^2-1=\left(n-1\right)\left(n+1\right)\)

\(\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}\)

Vậy .............

a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

Th1:  \(5x-1=0\Rightarrow5x=1\Rightarrow x=\frac{1}{5}\)

Th2: \(2x-\frac{1}{3}=0\Rightarrow2x=\frac{1}{3}\Rightarrow x=\frac{1}{6}\)

Vậy = 1/5 ; 1/6

b) \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)

\(\left|\frac{4}{5}-x\right|=\frac{1}{4}\)

Th1:

\(\frac{4}{5}-x=\frac{1}{4}\)

\(x=\frac{11}{20}\)

Th2:

\(\frac{4}{5}-x=-\frac{1}{4}\)

\(x=\frac{21}{20}\)

Vậy x= 11/20 và x= 21/20

c) \(\left|2\frac{1}{2}+x\right|-\left(-\frac{2}{3}\right)=3\)

\(\left|\frac{5}{2}+x\right|+\frac{2}{3}=3\)

\(\left|\frac{5}{2}+x\right|=\frac{7}{3}\)

Th1:

\(\frac{5}{2}+x=\frac{7}{3}\)

\(x=-\frac{1}{6}\)

Th2: 

\(\frac{5}{2}+x=-\frac{7}{3}\)

\(x=-\frac{29}{6}\)

Vậy x= -1/6 và x= -29/6

Ta có :-5x⁴< hoặc = 0(*)

           -9x²< hoặc = 0(**)

            -4<0(***)

TỪ (*);(**);(***) suy ra -5x⁴-9x²-4< hoặc = -4

Vậy đa thức N(x)=-5x⁴-9x²-4 là vô nghiệm (không có nghiệm)

Huỳnh Thị Thiên Kim: phân tích hằng đẳng thức

Chứng tỏ đa thức sau vô nghiệm :

Ta fai cho đa thức đó là : 0

Như sau :-5x^4-9x^2-4=0

Rồi tính như bài tim x bình thường

giúp mình với

a. \(\dfrac{x^2+2x+3}{x^2-x+1}=0\) ⇔x²+2x+3=0 ⇔x²+2x+1+2=0 ⇔(x+1)²+2=0

Vì (x+1)²+2>0 nên phương trình đã cho vô nghiệm.

b) \(\dfrac{x}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{x^2-4}\) ⇔\(\dfrac{x\left(x-2\right)+4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)

⇔\(x\left(x-2\right)+4\left(x+2\right)=4\) ⇔x²-2x+4x+8-4=0 ⇔x²+2x+4=0                ⇔x²+2x+1+3=0 ⇔(x+1)²+3=0

Vì (x+1)²+3>0 nên phương trình đã cho vô nghiệm.

1)

a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)

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

=>đpcm

b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)

=>đpcm

2,

a) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)

\(\Leftrightarrow50x=-100\)

\(\Leftrightarrow x=-2\)

b) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)

\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)

\(\Leftrightarrow-0,69x=0,138\)

\(\Leftrightarrow x=-0,2\)

Câu 1:

a)\(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^2-x+3\right)\)

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

\(=x^3+3\)(ko thể CM)

b)\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)

\(=-24\)(đpcm)

1 a ) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)

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

\(=3\)

=> đpcm

 b ) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)

\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)

\(=-24\)

=> đpcm

2 ) Tìm x :

a ) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-60x^2+15x=100\)

\(\Leftrightarrow50x=100\)

\(\Leftrightarrow x=2\)

b ) \(0,6c\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)

\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)

\(\Leftrightarrow-0,69x=0,138\)

\(\Leftrightarrow-0,2\)

\(N=-\left(5x^4+9x^2+4\right)=-\left(5x^4+5x^2+4x^2+4\right)=-\left(5x^2+4\right)\left(x^2+1\right)< 0\)

Do đó: Đa thức N(x) vô nghiệm