Question

3. Tìm x.

a) |x+1|+|x+5|=4

b) |2x-1|+|x-3y|=0

Answer 1

a) |x+1|+|x+5|=4

\(\Rightarrow x+1+x+5=\pm4\)

\(x+1+x+5=4\)

\(\Rightarrow x^2+1+5=4\)

\(x^2+6=4\)

\(x^2=4-6\)

\(\Rightarrow x^2=-2\)

\(x+1+x+5=-4\)

\(x^2+6=-4\)

\(x^2=-8\)

 

Answer 2

a) trường hợp 1:x\(\ge\)-1

x+1+x+5=4\(\Rightarrow2x+6=4\Rightarrow x=-1\)(TM)

TH2:\(-5\le x< -1\)

-x-1+x+5=4(phương trình vô nghiệm)

TH3:x<-5

-x-1-x-5=4\(\Rightarrow-2x-6=4\Rightarrow-5\)(KTM)

vậy x=-1

b)

 

 

 

 

 

 

Answer 3

b: Ta có: \(\left|2x-1\right|\ge0\forall x\)

\(\left|x-3y\right|\ge0\forall x,y\)

Do đó: \(\left|2x-1\right|+\left|x-3y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{1}{2};\dfrac{1}{6}\right)\)