\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\) 

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)

 \(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\) 

=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005

\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\Leftrightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}-\frac{x+3+2003}{2003}-\frac{x+4+2001}{2001}=0\)

x+1/x^2+x+1 -(x-1)/x^2+x+1=3/x(x^4+x^2+1)

đkxđ x khác 0

[(x+1)(x^2-x+1)-(x-1)(x^2+x+1)] /(x^2+x+1)(x^2-x+1)=3/x(x^4+x^2+1)

[(x^3+1)-(x^3-1)]/x^4+x^2+1=3/x(x^4+x^2+1)

nhân 2 vế pt cho x(x^4+x^2+1) ta được 

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

<=> 2x=3

<=>x=3/2 (thỏa)

S={3/2}

Đặt \(x^2+x+1=a\ne0vàx^2-x+1=b\ne0\)

\(\Rightarrow b-a=-2xvàb+a=2x^2+2\)

    và điều kiện \(x\ne0\)

thì  \(x\left(x^4+x^2+1\right)=xab\)

\(\Rightarrow PT\Leftrightarrow\frac{x+1}{a}-\frac{x-1}{b}=\frac{3}{xab}\)

              \(\Leftrightarrow\frac{bx\left(x+1\right)-ax\left(x-1\right)}{xab}=\frac{3}{xab}\)

             \(\Leftrightarrow bx^2+bx-ax^2+ax=3\)

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

             \(\Leftrightarrow2x-3=0\)

             \(\Leftrightarrow x=\frac{3}{2}\)(tm)

Vậy \(x=\frac{2}{3}\) là nghiệm của pt

Bạn tự tìm điều kiện xác định nhé :)

\(Q=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)

\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{3}{\sqrt{x}+3}:\frac{9-x+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}:\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}=\frac{3}{\sqrt{x}-2}\)

có 2 ý hả bạn

có 2 ý thì mik giải dc

Đề tớ gõ sai, Sr các cậu...

Đề đúng là :

\(\frac{x-3}{90}+\frac{x-2}{91}+\frac{x-1}{92}=3\)

Giúp tớ nhen...Giải chi tiết giùm nha...Thank you !!!

\(\left(\frac{x-3}{90}-1\right)+\left(\frac{x-2}{91}-1\right)+\left(\frac{x-1}{90}-1\right)=0\)

\(\Leftrightarrow\frac{x-93}{90}+\frac{x-93}{91}+\frac{x-93}{92}=0\)

\(\Leftrightarrow\left(x-93\right)\left(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\right)=0\)

mà \(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\ne0\)

\(\Leftrightarrow x-93=0\Leftrightarrow x=93\)

Vậy x=93

\(N=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\) \(=\frac{x-6\sqrt{x}-3\sqrt{x}+18-x}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}\) \(=\frac{18-9\sqrt{x}}{x-36}\)

\(2.x=\frac{1+2+3+...+9}{1-2+3-4+5-6+7-8+9}+\frac{25.150-60.5+20.75}{1+2+3+...+99}\)

\(2.x=\frac{\left(9+1\right).9:2}{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+9}+\frac{2.3.5^2.\left(5^2-2+2.5\right)}{\left(1+99\right).99:2}\)

\(2.x=\frac{45}{\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+9}+\frac{2.3.5^2.33}{100.99.\frac{1}{2}}\)

\(2x=\frac{45}{5}+\frac{50.99}{50.2.99.\frac{1}{2}}=9+\frac{1}{2.\frac{1}{2}}=9+1=10\)

=> 2x = 10

x = 5

\(\frac{x-2015}{2}+\frac{x-2016}{3}=\frac{x-2017}{4}+\frac{x-2018}{5}\)

\(=\frac{x-2015}{2}+1+\frac{x-2016}{3}+1=\frac{x-2017}{4}+1+\frac{x-2018}{5}+1\)

\(\frac{x-2013}{2}+\frac{x-2013}{3}=\frac{x-2013}{4}+\frac{x-2013}{5}\)

\(\frac{x-2013}{2}+\frac{x-2013}{3}-\frac{x-2013}{4}-\frac{x-2013}{5}=0\)

\(\left(x-2013\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

vì \(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)nên \(x-2013=0\)

x = 2013

phan van hieu tuyet

x-2107/4 à đề kiểu gì vậy