Question

Tìm x : a) \(\left|5x-4\right|=\left|x+2\right|\)

b) \(5-\left|x-2\right|=3-\left(2x+1\right)\)

Answer 1

Với những bài thế này thì phải chia trường hợp để phá ngoặc.

TH1 : \(x< -2;\)có:

\(\Rightarrow-\left(5x-4\right)=-\left(x+2\right)\)

\(4-5x=-x-2\)

\(6=-4x\Rightarrow x=-\frac{3}{2}>-2\)( Không thỏa mãn )

TH2 : \(-2\le x< \frac{4}{5};\)ta có :

\(-\left(5x-4\right)=x+2\)

\(4-5x=x+2\)

\(2=6x\)

\(x=\frac{1}{3}\) ( thỏa mãn)

TH3 : \(x\ge\frac{4}{5};\)có :

\(5x-4=x+2\)

\(4x=6\)

\(x=\frac{3}{2}\)(thỏa mãn )

Vậy \(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=\frac{3}{2}\end{array}\right.\)

 

Answer 2

Dũng sai

Answer 3

a) |5x-4| = |x+2|

<=> \(\left[\begin{array}{nghiempt}5x-4=x+2\\5x-4=-\left(x+2\right)\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}4x=6\\5x-4=-x-2\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\4x=2\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{1}{2}\end{array}\right.\)

( không chắc đâu, t dốt qt chuyển vế lắm :v )

Answer 4

Mình hiểu rồi làm lại nè:

|5x-4| = |x+2|

<=> \(\left[\begin{array}{nghiempt}5x-4=x+2\\5x-4=-x-2\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}4x=6\\6x=2\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{1}{3}\end{array}\right.\)