ĐKXĐ: ...

Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}=a^2\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)

\(\Rightarrow\frac{x^2}{3}+\frac{48}{x^2}=3a^2+8\)

\(3a^2+8=10a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)

ĐKXĐ: ...

Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow a^2=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)

\(a^2+\frac{8}{3}=\frac{10}{3}a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)

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

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

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

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

\(\Rightarrow x=\pm1\)

a,Bạn xét 3 th

th1: x>=-1

th2: 1>x>-1

th3:x<=1

rồi trong từng th bạn bỏ dấu gttd và giải

b, \(\frac{x^2}{3}+\frac{48}{x^2}=10\left(\frac{x}{3}-\frac{4}{x}\right)\)

tương đương \(x^2+\frac{144}{x^2}=10\left(x-\frac{12}{x}\right)\)(nhân cả 2 vế với 3)

tương đương \(\left(x-\frac{12}{x}\right)^2+24-10\left(x-\frac{12}{x}\right)\)=0

đặt (x-12/x)=a

khi đó a^2-10a+24=0

giải a rồi tìm x thôi 

c, đặt \(\sqrt[3]{x}\)=a

khi đó ta có 2a^2-5a=3

giải a rồi tìm x thôi

Chúc bạn học tốt!

Ta có : x^4+x^2+1

=x^4+x+x^2-x+1

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

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

Suy ra ta có phương trình :

  X  -1     _    X  + 1   =       10                     

X^2-X +1    X^2+X +1     X(X^2-X+1)(X^2+X+1)

<=>  X^3 - 1 - (  X^3 + 1)       =        10                   

       (X^2-X+1)(X^2+X+1)            X(X^2-X+1)(X^2+X+1)

<=>         -2X                        =           10                  

         X(X^2-X+1)(X^2+X+1)         X(X^2-X+1)(X^2+X+1)

<=> -2X=10

<=>x =-5

vậy x=-5

Ta có :

\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)

\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)

\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)

\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)

Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)

\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài 

Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)

Phương trình đầu bài tương đương với 
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)

Vậy phương trình có nghiệm duy nhất là x=-100

<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)

=> x+100=0 => x= -100

vay pt co nghiem \(x=-100\)

Ta thấy:\(\frac{x+43}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(2\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{x+52}{48}\)\(+\)\(2\)

\(\Rightarrow\)\(\frac{x+43}{57}\)\(+\)\(\frac{57}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(\frac{54}{54}\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{51}{51}\)\(+\)\(\frac{x+48}{52}\)\(+\)\(\frac{52}{52}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}\)\(+\)\(\frac{x+100}{54}\)\(=\)\(\frac{x+100}{51}\)\(+\)\(\frac{x+100}{52}\)

\(\Leftrightarrow\)\((\)\(x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\()\)\(=\)\((x+100)\)\((\frac{1}{52}\)\(+\)\(\frac{1}{51})\)

\(\Leftrightarrow\)\((x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)\()\)\(=\)\(0\)\((1)\)

Ta thấy: \(\frac{1}{57}\)< \(\frac{1}{52}\)

          \(\frac{1}{54}\)<\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)< \(\frac{1}{52}\)\(+\)\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)< 0 \((2)\)

Từ \((1)\)và \(\left(2\right)\)\(\Rightarrow\)\(x+100\)\(=0\)

\(\Leftrightarrow x=-100\)

Vậy phương trình có tập nghiệm \(x=-100\)

Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{9b+1}{4}\\\frac{2\cdot\left(9b+1\right)}{4}-9b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{-1}{9}\\a=\frac{9b+1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=\frac{-1}{9}\end{matrix}\right.\)

Pt vô nghiệm.

Do \(x^2\ge0\Rightarrow x^2+1\ge1\Rightarrow\frac{1}{x^2+1}>0.\)

Tương tự \(\frac{1}{x^2+2};\frac{1}{x^2+3};\frac{1}{x^2}+4>0\)

=> Phương trình vô nghiệm