Question

Tìm x

a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)

b,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

 

Answer 1

a) Dễ thấy VT > 0;mà VT=VP

=>VP > 0 => 4x > 0=> x > 0

=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)

=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)

\(=>3x+1=4x=>x=1\)

Answer 2

a)  Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )

Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)

=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)

<=>x=1

Vậy x=1

b)Điều kiện: \(x\ne-3;-10;-21;-34\)

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

=>x+34-x-3=x

<=>x=31 (nhận)

Vậy x=31

Answer 3

a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)

Ta có: \(\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{3}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\end{cases}\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|\ge0\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}\)

Khi đó, ta có: \(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)

\(\Rightarrow3x+1=4x\)

\(\Rightarrow x=1\)

b) Từ đề suy ra:

\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{x+34}{\left(x+3\right)\left(x+34\right)}-\frac{x+3}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow x=31\)

Answer 4

b) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(=>\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(=>\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}=>\frac{x+34-x-3}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(=>\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}=>x=31\)