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1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009

x+(-31/12)^2=(49/12)^2-x
x+x=(49/12)^2-(-31/12)^2
tính x
từ x tìm ra y
b)x(x-y):[y(x-y)]=3/10:(-3/50)=...
=>x/y=... =>x=...;y=...

a, \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{5x}{15}=\frac{2y}{8}=\frac{5x-2y}{15-8}=\frac{28}{7}=4\)
=> x = 4.3 = 12
y = 4.4 = 16
b, \(x:2=y:\left(-5\right)\Rightarrow\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{-7}{7}=-1\)
=> x = (-1).2 = -2
y = (-1)(-5) = 5
c, \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-10}=\frac{10}{10}=1\)
=> x = 8
y =12
z = 15

|x^4+x^2+1| = |x^2+x+1|
=> x^4+x^2+1 = x^2+x+1
=> x^4+x^2+1-x^2-x-1 = 0
=> x^4-x = 0
=> x.(x^3-1) = 0
=> x=0 hoặc x^3-1=0
=> x=0 hoặc x=1
Vậy .........

\(x^2-x=x-2\)
\(x^2-x-x+2=0\)
\(x^2-2x+2=0\)
\(x^2-\left(2x-2\right)=0\)
\(x^2-2\left(x-1\right)=0\)
\(x^2-1-2\left(x-1\right)+1=0\)
\(\left(x-1\right)\left(x+1\right)-2\left(x-1\right)+1=0\)
\(\left(x-1\right)\left(x+1-2\right)+1=0\)
\(\left(x-1\right)\left(x-1\right)=-1\)
\(\left(x-1\right)^2=-1\)(vô lí)
Vậy không tìm được x

1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)

1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)

Ta có :
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
Theo tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}=\frac{x-1+x-2}{x+2+x+3}\)
\(=\frac{2x-3}{2x+5}=\frac{2x+5-8}{2x+5}=1+\frac{8}{2x+5}\)
\(\Rightarrow2x+5\)thuộc \(Ư\left(8\right)=\left(1;-1;2;-2;4;-4;8;-8\right)\)
\(\Rightarrow2x\)thuộc \(\left(-4;-6;-3;-7;-1;-9\right)\)
\(\Rightarrow x\)thuộc \(\left(-2;-3;\frac{-3}{2};-\frac{7}{2};\frac{-1}{2};\frac{-9}{2}\right)\)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Leftrightarrow(x-1)(x+3)=(x+2)(x-2)\)
\(\Leftrightarrow3x-3=x^2-4\)
\(\Leftrightarrow-3+4=x^2-3x\)
\(\Leftrightarrow1=x(x-3)\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\x-3=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=4\end{cases}}\)
\(\left|x-2\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x\ge2\\0\le x< 2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=x\\2-x=x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2=0\left(\text{vô nghiệm}\right)\\x=1\end{cases}}\)
Trả lời:
| x - 2 | = x
=> x - 2 = x hoặc x - 2 = - x
=> x - x = 2 hoặc x + x = 2
=> 0x = 2 ( vô lí ) hoặc 2x = 2
=> x = 1
Vậy x = 1