\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

a) => 4x + 2/3 = 0 hoặc 2/3x - 1 =0 

4x= -2/3 hoặc 2/3x= 1

x = -2/3 . 1/4 hoặc x = 1.3/2

x = -1/6 hoặc x = 3/2 

b) x+2 / x -1 = 5/2 

=> 2(x+2) = 5(x-1)

2x + 4 = 5x - 5

5x - 2x= 4+5

3x = 9

=> x= 3

a) (4x+\(\frac{2}{3}\)) . ( \(\frac{2}{3}\)x-1)=0

\(\Rightarrow\)\(\orbr{\begin{cases}4x+\frac{2}{3}=0\\\frac{2}{3}x-1=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)........

Tới đây bn tự giải nha

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+99\right|=100x\)

\(\left|x+1\right|\ge0;\left|x+2\right|\ge0;...;\left|x+99\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+...+x+99=100x\)

\(\Rightarrow99x+1+2+3+...+99=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow-x=-4950\)

\(\Rightarrow x=4950\)

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{49\cdot50}\right|=50x\)

\(\left|x+\frac{1}{1\cdot2}\right|\ge0;\left|x+\frac{1}{2\cdot3}\right|\ge0;...;\left|x+\frac{1}{49\cdot50}\right|\ge0\)

\(\Rightarrow50x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{49\cdot50}\)

\(\Rightarrow49x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

tu lam 

\(a;\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+..............+\left|x+99\right|=100x^{\left(1\right)}\)

Ta có \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+3\right|\ge0;.............;\left|x+99\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\).Từ (1) \(\Rightarrow x+1+x+2+x+3+..................+x+99=100x\)

\(\Rightarrow\left(x+x+x+........+x\right)+\left(1+2+3+..........+99\right)=100x\)

\(\Rightarrow99x+4950=100x\)

\(\Rightarrow x=4950\)(t/m đk x > =  0)

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+.........+\left|x+\frac{1}{49.50}\right|=50x^{(∗)}\)

\(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;............;\left|x+\frac{1}{49.50}\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow50x\ge0\Rightarrow x\ge0\)

Với x > = 0 .Từ (*) \(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+............+x+\frac{1}{49.50}=50x\)

\(\Rightarrow\left(x+x+x+.......+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49.50}\right)=50x\)

\(\Rightarrow49x+\left(1-\frac{1}{50}\right)=50x\)

\(\Rightarrow49x+\frac{49}{50}=50x\)

\(\Rightarrow x=\frac{49}{50}\)(t/m đk \(x\ge0\))

\(\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y}{z}\Rightarrow k=2\Rightarrow x=y=z=1\)

A=6

\(\frac{x-y-z}{x}=1-\frac{y+z}{x}\) tương tự con khác

=> x=y=z

=> A=6