\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)

\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)

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

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

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

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

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

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

\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=2\\x=-\frac{2}{3}\end{cases}}\)

\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\)

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

\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)

\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)

\(\Leftrightarrow3x^2-7x^2+4=0\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)

vậy:...

ĐK \(x\ne\left\{-1;-\frac{1}{2}\right\}\)

Phương trình \(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1=\frac{-x^2+5x-1}{2x+1}-1\)\(\Leftrightarrow\frac{x^2-4x+1+x+1}{x+1}=\frac{-x^2+5x-1-2x-1}{2x+1}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+2=0\\\frac{1}{x+1}+\frac{1}{2x+1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)\left(x-2\right)=0\\\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1;x=2\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1;x=2\\x=-\frac{2}{3}\end{cases}}\left(tm\right)}\)

Vậy hệ có 3 nghiệm \(x=1;x=2;x=-\frac{2}{3}\)

\(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1=-\frac{x^2-5x+1}{2x+1}-1.DKXD:x\ne-1;x\ne-\frac{1}{2}\)

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

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

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

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

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(n\right)\)

\(hay:x-2=0\Leftrightarrow x=2\left(n\right)\)

\(hay:\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\left(n\right)\)

\(V...S=\left\{1:2:-\frac{2}{3}\right\}\)

Đặt \(x+\frac{1}{x}=t\Rightarrow\left(x+\frac{1}{x}\right)^2=t^2\Leftrightarrow x^2+\frac{1}{x^2}=t^2-2\)

Khi đó phương trình đã cho 

\(\Leftrightarrow2t^2+\left(t^2-2\right)^2-t^2\left(t^2-2\right)=4-4x+x^2\)

\(\Leftrightarrow2t^2+t^4-4t^2+4-t^4+2t^2=x^2-4x+4\)

\(\Leftrightarrow4=x^2-4x+4\)

\(\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Mà ĐKXĐ của phương trình là \(x\ne0\)

Tập nghiệm của pt là \(S=\left\{4\right\}\)

Đặt \(x+\frac{1}{x}=a\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow x^2+\frac{1}{x^2}+2=a^2\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)

Có \(2a^2+\left(a^2-2\right)^2-a^2\left(a^2-2\right)=\left(2-x\right)^2\)

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

\(\Leftrightarrow4=\left(2-x\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2-x=4\\2-x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

Vậy \(S=\left(-2;6\right)\)

Tại sao \(\left(x^2+\frac{1}{x^2}\right)=t^2-2\) thế

hic, mk chx học

A=1/(x-2)(x-3) + 1/(x-3)(x-4) + 1/(x-4)(x-5) + 1/(x-5)(x-6)=1/8 (ĐKXĐ: x#2,x#3,x#4,x#5,x#6)

A= 1/x-2 -1/x-3 + 1/x-3 -1/x-4 .....-1/x-6=1/8

=>1/x-2 -1/x-6=1/8

=>8(x-6)-8(x-2)=(x-2)(x-6)

=> 8x-48-8x+16=x^2-8x+12

=> x^2-8x-20=0

=> (x-10)(x+2)=0 => x=10,x=-2 thuộc ĐKXĐ

Có cần thế ko ạ ??? Shinichi

Điều kiện xác định \(\hept{\begin{cases}x\ne2\\x\ne\\x\ne4\end{cases}3}\)

                              \(\hept{\begin{cases}x\ne5\\x\ne6\end{cases}}\)

Ta có : \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(x^2-11+30=\left(x-5\right)\left(x-6\right)\)

Phương trình đã tương đương với 

\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\)

\(\Leftrightarrow x^2-8x-20=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)

\(x-10=0\Leftrightarrow x=10\)

hoặc 

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

\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)thỏa mãn điều kiện phương trình 

Phương trình có nghiệm \(x=10;x=-2\)

Ai làm đc câu nào thì làm giúp mình với ạ, cảm ơn trc:(((

\(1,3x-5x+5=-8\)

\(\Leftrightarrow-2x+5+8=0\)

\(\Leftrightarrow-2x=-13\)

\(\Leftrightarrow x=\frac{13}{2}\)

dell biết

Lời giải:

a) $2x-1=x-3x^2$

$\Leftrightarrow 3x^2+x-1=0$

$\Leftrightarrow 36x^2+12x-12=0$

$\Leftrightarrow (6x+1)^2=13$

$\Rightarrow 6x+1=\pm \sqrt{13}$

$\Rightarrow x=\frac{1\pm \sqrt{13}}{6}$

b) Bạn xem lại xem có nhầm dấu không?