Question

[cà thái thành]

tìm x trong các đẳng thức:

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

c) \(\left|x-1\right|+3x=11\)                  d)\(\left|5x-3\right|-x=7\)

Answer 1

[.]

a) \(\left|2x-3\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

Vậy : \(x\in\left\{4,-1\right\}\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2x-3\\2x-1=3-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-2x=-3+1\\2x+2x=3+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0=-2\\4x=4\end{cases}}\)

\(\Rightarrow x=1\)

Vậy : \(x=1\)

c) \(\left|x-1\right|+3x=11\)

\(\Leftrightarrow\left|x-1\right|=11-3x\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=11-3x\\x-1=3x-11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3x=11+1\\3x-x=-1+11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=12\\2x=10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

Vậy : \(x\in\left\{3,5\right\}\)

d) \(\left|5x-3\right|-x=7\)

\(\Leftrightarrow\left|5x-3\right|=7+x\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{4}=\frac{5}{2}\\x=-\frac{4}{6}=-\frac{2}{3}\end{cases}}\)

Vậy : \(x\in\left\{\frac{5}{2},-\frac{2}{3}\right\}\)

Answer 2

[Hồ Hoàng Trúc Vân]

a,\(|2x-3|=5\)

\(\Rightarrow2x-3=5\)hoặc\(2x-3=-5\)

\(2x=5+3\)hoặc\(2x=-5+3\)

\(2x=8\)hoặc\(2x=-2\)

\(\Rightarrow x=\frac{8}{2}=4\)hoặc\(x=-\frac{2}{2}=-1\)

Vậy.......

b,Sai vế phải r bn ơi

c,\(|x-1|+3x=11\)

\(|x-1|=11-3x\)

\(\Rightarrow x-1=+\left(11-3x\right)\)hoặc\(\Rightarrow x-1=-\left(11-3x\right)\)

\(x-1=11-3x\)hoặc\(x-1=-11+3x\)

\(x+3x=11+1\)hoặc\(x-3x=-11+1\)

\(4x=12\)hoặc\(-2x=-10\)

\(\Rightarrow x=\frac{12}{4}=3\)hoặc\(x=\frac{-10}{-2}=5\)

Vậy.....

d,\(|5x-3|-x=7\)

\(|5x-3|=7+x\)

\(\Rightarrow5x-3=+\left(7+x\right)\)hoặc\(5x-3=-\left(7+x\right)\)

\(5x-3=7+x\)hoặc\(5x-3=-7-x\)

\(5x-x=7+3\)hoặc\(5x+x=-7+3\)

\(4x=10\)hoặc\(6x=-4\)

\(\Rightarrow x=\frac{10}{4}=2,5\)hoặc\(x=-\frac{4}{6}=-\frac{2}{3}\)

Vậy......