\(x\ne\left\{-4;-3;-2;-1\right\}\)

\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)

\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)

\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)

\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)

\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{-2}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+2\right)\left(x+4\right)}\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)

ĐK \(x\ne0,x\ne-1\)

Ta có \(\frac{x^2-4+\frac{1}{x^2}}{x+\frac{1}{x}}+x^2+3+\frac{1}{x^2}=4\)

Đặt \(x+\frac{1}{x}=a\)=> \(x^2+\frac{1}{x^2}=a^2-2\)

=> \(\frac{a^2-6}{a}+a^2-3=0\)

<=> \(a^3+a^2-3a-6=0\)=> \(\left(a-2\right)\left(a^2+3a+3\right)=0\)

                                                          => a=2

=> \(x+\frac{1}{x}=2\)=> \(x^2+1=2x\)=> x=1 (thỏa mãn ĐKXĐ)

Vậy \(x=1\)

\(ĐKXĐ:x\ne0\)

\(PT\Leftrightarrow\frac{x^7-x^6+4x^5-4x^4+4x^3+x^2+x}{x^3\left(x^2+1\right)}=4\)

\(\Leftrightarrow\frac{x^6+x^5-4x^3+x+1+4x^2\left(x^2+1\right)}{x^2\left(x^2+1\right)}=4\)

\(\Leftrightarrow\frac{x^6+x^5-4x^3+x+1}{x^2\left(x^2+1\right)}=0\)

\(\Leftrightarrow x^6+x^5-4x^3+x+1=0\)

\(\Leftrightarrow x^6-x^5+2x^5-2x^4+2x^4-2x^3-2x^3+2x^2-2x^2+2x-x+1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5+2x^4+2x^3-2x^2-2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^4+3x^3+5x^2+3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^4+3x^3+5x^2+3x+1\right)=0\)

Vì \(x^4+3x^3+5x^2+3x+1\ne0\)nên

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

Vậy tập nghiệm của pt là \(S=\left\{1\right\}\)

Trần Phúc Khang x³ + x = x(x² + 1) nên x = -1 vẫn ok mà . Còn lại thì y như cách em làm ban sáng.

ĐK: x>0

Đặt a=1/x ta được: a>0

\(a+\frac{1}{3}=\sqrt{\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}}\)

\(\Leftrightarrow a^2+\frac{1}{9}+\frac{2}{3}a=\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{2}{3}a=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a.\left(a+\frac{2}{3}\right)=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a+\frac{2}{3}=\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{4}{9}+\frac{4}{3}a=\frac{4}{9}+2a^2\)

<=>\(a^2-\frac{4}{3}a=0\Leftrightarrow a=0\left(loại\right);a=\frac{4}{3}\)

<=>\(x=\frac{3}{4}\)(loại -3/2)

Vậy x=3/4

\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\)                              \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)

\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)

\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)

\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)

Vậy pt vô nghiệm

= $\frac{3+x}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}$3+x3x =√19 ‍+1x √49 +2x2 

a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

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

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

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

Câu 1 :

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

=> \(3x-3+2=3x-1\)

=> \(3x-3+2-3x+1=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

=> \(-19x+114=0\)

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)

⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0

⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0

Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0

⇔x+2014=0

⇔x=-2014

Vậy tập nghiệm của phương trình đã cho là:S={-2014}

2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)

⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)

⇒9x+6+6x+18=4x-4+x+1

⇒15x+24=5x-3

⇒15x-5x=-3-24

⇒10x=-27

⇒ x=-\(\frac{27}{10}\)

Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}

\(3.\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0ĐKXĐ:x\ne\pm1\)

\(\frac{1+x}{x-1}-\frac{x-1}{1+x}+\frac{x^2+3}{\left(1+x\right)\left(1-x\right)}=0\)

\(-3+7x-5x^2+x^3=0\)

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

\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x=3